Prep: Data Types

posted Nov 10, 2011, 1:08 AM by Neil Mathew   [ updated Nov 10, 2011, 1:27 AM ]


Already briefly discussed: Datatypes

Note: Noticed in oracle:

NUMBER ( 5, 2 ) is a floating type number which allows a column to have a 5digit number with two decimal places. That is, only 3 non-decimal places are allowed. eg: 300.02

I had some errors arising because I entered less digits in the decimal places or too many overall. Just a note.


CHAR(size)Fixed length character data of length size bytes. This should be used for fixed length data. Such as codes A100, B102...
VARCHAR2(size)Variable length character string having maximum length sizebytes.
You must specify size
NUMBER(p,s)Number having precision p and scale s.

LONGCharacter data of variable length (A bigger version the VARCHAR2 datatype)
up to 2 Gbytes of data

RAW(size)Raw binary data of length size bytes
used for byte-oriented data (to store character strings, floating point data, and binary data such as graphics images and digitized sound )
You must specify size for a RAW value.
LONG RAWRaw binary data of variable length. (not intrepreted by PL/SQL)

NCLOBNational Character Large Object

BLOBBinary Large Object
like graphics, sound,etc

CLOBCharacter Large Object
like text,memos, etc

BFILEpointer to binary file on disk

DATEValid date range


The NUMBER datatype 

Stores zero, positive, and negative numbers, fixed or floating-point numbers

Fixed-point NUMBER
NUMBER(p,s
precision p = length of the number in digits 
scale s = places after the decimal point, or (for negative scale values) significant places before the decimal point. 

Integer NUMBER
NUMBER(p
This is a fixed-point number with precision p and scale 0. Equivalent to NUMBER(p,0) 

Floating-Point NUMBER
NUMBER 
floating-point number with decimal precision 38



Prep: Constraints

posted Nov 10, 2011, 12:47 AM by Neil Mathew   [ updated Nov 10, 2011, 12:47 AM ]

SQL Constraints

Constraints are used to limit the type of data that can go into a table.

Constraints can be specified when a table is created (with the CREATE TABLE statement) or after the table is created (with the ALTER TABLE statement).


(Note that I am aware that most of these examples are not suited for Oracle.)


We will focus on the following constraints:

  • NOT NULL

    The NOT NULL constraint enforces a column to NOT accept NULL values.

    The NOT NULL constraint enforces a field to always contain a value. This means that you cannot insert a new record, or update a record without adding a value to this field.


  • UNIQUE

    SQL UNIQUE Constraint

    The UNIQUE constraint uniquely identifies each record in a database table.

    The UNIQUE and PRIMARY KEY constraints both provide a guarantee for uniqueness for a column or set of columns.

    A PRIMARY KEY constraint automatically has a UNIQUE constraint defined on it.

    Note that you can have many UNIQUE constraints per table, but only one PRIMARY KEY constraint per table.


    (to add UNIQUE CONSTRAINT using Alter table)

    ALTER TABLE Persons ADD CONSTRAINT uc_PersonID UNIQUE (P_Id,LastName)


  • PRIMARY KEY

    The PRIMARY KEY constraint uniquely identifies each record in a database table.

    Primary keys must contain unique values.

    A primary key column cannot contain NULL values.

    Each table should have a primary key, and each table can have only ONE primary key.



  • FOREIGN KEY

    A FOREIGN KEY in one table points to a PRIMARY KEY in another table.

    Let's illustrate the foreign key with an example. Look at the following two tables:

    The "Persons" table:

    P_IdLastNameFirstNameAddressCity
    1HansenOlaTimoteivn 10Sandnes
    2SvendsonToveBorgvn 23Sandnes
    3PettersenKariStorgt 20Stavanger

    The "Orders" table:

    O_IdOrderNoP_Id
    1778953
    2446783
    3224562
    4245621

    Note that the "P_Id" column in the "Orders" table points to the "P_Id" column in the "Persons" table.

    The "P_Id" column in the "Persons" table is the PRIMARY KEY in the "Persons" table.

    The "P_Id" column in the "Orders" table is a FOREIGN KEY in the "Orders" table.

    The FOREIGN KEY constraint is used to prevent actions that would destroy links between tables.

    The FOREIGN KEY constraint also prevents that invalid data form being inserted into the foreign key column, because it has to be one of the values contained in the table it points to.



    CREATE TABLE Orders

    (

    O_Id int NOT NULL,

    OrderNo int NOT NULL,

    P_Id int,

    PRIMARY KEY (O_Id),

    FOREIGN KEY (P_Id) REFERENCES Persons(P_Id)

    )


    CREATE TABLE Orders
    (
    O_Id int NOT NULL PRIMARY KEY,
    OrderNo int NOT NULL,
    P_Id int FOREIGN KEY REFERENCES Persons(P_Id)
    )


    CREATE TABLE Orders
    (
    O_Id int NOT NULL,
    OrderNo int NOT NULL,
    P_Id int,
    PRIMARY KEY (O_Id),
    CONSTRAINT fk_PerOrders FOREIGN KEY (P_Id)
    REF
    ERENCES Persons(P_Id)
    )


    ALTER TABLE Orders

    ADD FOREIGN KEY (P_Id)

    REFERENCES Persons(P_Id)


    To allow naming of a FOREIGN KEY constraint, and for defining a FOREIGN KEY constraint on multiple columns, use the following SQL syntax:


    ALTER TABLE Orders

    ADD CONSTRAINT fk_PerOrders

    FOREIGN KEY (P_Id)

    REFERENCES Persons(P_Id)


  • CHECK
The CHECK constraint is used to limit the value range that can be placed in a column.

If you define a CHECK constraint on a single column it allows only certain values for this column.

If you define a CHECK constraint on a table it can limit the values in certain columns based on values in other columns in the row.


CREATE TABLE Persons
(
P_Id int NOT NULL,
LastName varchar(255) NOT NULL,
FirstName varchar(255),
Address varchar(255),
City varchar(255),
CHECK (P_Id>0)
);


CREATE TABLE Persons
(
P_Id int NOT NULL CHECK (P_Id>0),
LastName varchar(255) NOT NULL,
FirstName varchar(255),
Address varchar(255),
City varchar(255)
)

The CHECK constraint is used to limit the value range that can be placed in a column.
If you define a CHECK constraint on a single column it allows only certain values for this column.If you define a CHECK constraint on a table it can limit the values in certain columns based on values in other columns in the row.

  • DEFAULT

The DEFAULT constraint is used to insert a default value into a column.

The default value will be added to all new records, if no other value is specified.


CREATE TABLE Persons
(
P_Id int NOT NULL,
LastName varchar(255) NOT NULL,
FirstName varchar(255),
Address varchar(255),
City varchar(255) DEFAULT 'Sandnes'
)

The DEFAULT constraint can also be used to insert system values, by using functions like GETDATE():

CREATE TABLE Orders
(
O_Id int NOT NULL,
OrderNo int NOT NULL,
P_Id int,
OrderDate date DEFAULT GETDATE()
)



Prep: Name the DDL, DML, DCL, TCL commands.

posted Nov 10, 2011, 12:06 AM by Neil Mathew   [ updated Nov 10, 2011, 12:06 AM ]

DDL


Data Definition Language (DDL) statements are used to define the database structure or schema. Some examples:
  • CREATE - to create objects in the database
  • ALTER - alters the structure of the database
  • DROP - delete objects from the database
  • TRUNCATE - remove all records from a table, including all spaces allocated for the records are removed
  • COMMENT - add comments to the data dictionary
  • RENAME - rename an object

DML


Data Manipulation Language (DML) statements are used for managing data within schema objects. Some examples:
  • SELECT - retrieve data from the a database
  • INSERT - insert data into a table
  • UPDATE - updates existing data within a table
  • DELETE - deletes all records from a table, the space for the records remain
  • MERGE - UPSERT operation (insert or update)
  • CALL - call a PL/SQL or Java subprogram
  • EXPLAIN PLAN - explain access path to data
  • LOCK TABLE - control concurrency

DCL


Data Control Language (DCL) statements. Some examples:
  • GRANT - gives user's access privileges to database
  • REVOKE - withdraw access privileges given with the GRANT command

TCL


Transaction Control (TCL) statements are used to manage the changes made by DML statements. It allows statements to be grouped together into logical transactions.
  • COMMIT - save work done
  • SAVEPOINT - identify a point in a transaction to which you can later roll back
  • ROLLBACK - restore database to original since the last COMMIT
  • SET TRANSACTION - Change transaction options like isolation level and what rollback segment to use

Prep: DDL Commands

posted Nov 9, 2011, 11:53 PM by Neil Mathew   [ updated Nov 10, 2011, 12:27 AM ]


An observation on my side.
Something that tended to confuse me was the usage of the word "table" in SQL syntax.

For example,
CREATE TABLE EMPLOYEE ( ....
ALTER TABLE EMPLOYEE ....

INSERT INTO EMPLOYEE VALUES ( ...
DELETE FROM EMPLOYEE ...

The question simple. The answer, even more so. The word 'TABLE' appears only in DDL commands. And the reason for specifying the 'TABLE' is because views too can be created using the same commands.

Examples: 
CREATE VIEW SOME_EMPLOYEES ...



DDL stands for Data Definition Language. 
It is the part of SQL programming language that deals with the construction and alteration of database structures like tables, views, and further the entities inside these tables like columns. It may be used to set the properties of columns as well. 

The three popular commands used in DDL are: 


Create - Used to create tables, views, and also used to create functions, stored procedures, triggers, indexes etc.

-- An example of Create command below
CREATE TABLE t_students ( 
stud_id NUMBER(10) PRIMARY KEY,
first_name VARCHAR2(20) NULL,
last_name VARCHAR2(20) NOT NULL,
dateofbirth DATE NULL);




Drop - Used to totally eliminate a table, view, index from a database - which means that the records as well as the total structure is eliminated from the database.

-- An example of Drop command below
DROP TABLE t_students;




Alter - Used to alter or in other words, change the structure of a table, view, index. This is particularly used when there is a scenario wherein the properties of fields inside a table, view, index are supposed to be updated.

-- An example of Alter command below

ALTER TABLE t_students ADD address VARCHAR2(200); -- Adds a column (^^ column NOT specified)

ALTER TABLE t_students DROP COLUMN dateofbirth; -- Drops a column (^^specify column)

ALTER TABLE t_students MODIFY COLUMN address VARCHAR2(100); -- Modifies a column 

ALTER TABLE t_students RENAME COLUMN supplier_name to sname; --Rename a column

-- You can also modify multiple columns using a single modify clause 

ALTER TABLE t_students MODIFY {
COLUMN address VARCHAR2(100)
COLUMN first_name VARCHAR2(50)
COLUMN last_name VARCHAR2(50)
}


-- You can rename a table using ALTER TABLE command instead of RENAME

Syntax:

ALTER TABLE table_name
 RENAME TO new_table_name;

For example:

ALTER TABLE suppliers
 RENAME TO vendors;


-- You can also add constraints like NOT NULL using the Modify statement 

ALTER TABLE t_students Modify 
{ first_name VARCHAR2(50) NOT NULL };   

or using ADD CONSTRAINT statement


ALTER TABLE t_students ADD CONSTRAINT to_nonull
            NOT NULL first_name;

Syntax: 
ALTER TABLE Employee ADD CONSTRAINT <constraint name> <constraint type> <column name>


-- Adding Referential Integrity

ALTER TABLE <table name> ADD CONSTRAINT <constraint name>
            FOREIGN KEY deptno )
            REFERENCES <parent table name> ( <primary key column> );

ALTER TABLE Employee ADD CONSTRAINT Ref_dept
            FOREIGN KEY (deptno)
            REFERENCES DEPT (deptno);



Rename - can be used to rename an object like a view or table

RENAME EMP_NM TO EMP_MW;



Others:

  • TRUNCATE - remove all records from a table, including all spaces allocated for the records are removed
  • COMMENT - add comments to the data dictionary

MySQL: 3 Cheat Sheet

posted Sep 3, 2011, 1:40 PM by Neil Mathew   [ updated Sep 3, 2011, 2:13 PM ]

 [ From BCA HUB ]


1) Display the details of all employees

SQL>Select * from emp;

 

2) Display the depart information from department table

SQL>select * from dept;

 

3) Display the name and job for all theemployees

SQL>select ename,job from emp;

 

4) Display the name and salary  for all the employees

SQL>select ename,sal from emp;

 

5) Display the employee no and totalsalary  for all the employees

SQL>select empno,ename,sal,comm, sal+nvl(comm,0) as”total  salary” from emp;

 

6) Display the employee name and annual salary for all employees.

SQL>select ename, 12*(sal+nvl(comm,0)) as “annual Sal” from emp;

 

7) Display the names of all the employees who are working in depart number 10.

SQL>select emame from emp where deptno=10;

 

8 ) Display the names of all the employees who are working as clerks and drawing a salary more than 3000.

SQL>select ename from emp where job=’CLERK’ and sal>3000;

 

9) Display the employee number and name  who are earning comm.

SQL>select empno,ename from emp where comm is not null;

 

10) Display the employee number and name  who do not earn any comm.

SQL>select empno,ename from emp where comm is null;

 

11) Display the names of employees who are working as clerks,salesman or analyst and drawing a salary more than 3000.

SQL>select ename  from emp where job=’CLERK’ OR JOB=’SALESMAN’ OR JOB=’ANALYST’ AND SAL>3000;

 

12) Display the names of the employees who are working in the company for the past 5 years;

SQL>select ename  from emp where to_char(sysdate,’YYYY’)-to_char(hiredate,’YYYY’)>=5;

 

13) Display the list of employees who have joined the company before 30-JUN-90 or after 31-DEC-90.

SQl> select ename from emp where hiredate < ’30-JUN-1990′ or hiredate >’31-DEC-90′;

 

14) Display current Date.

SQL>select sysdate from dual;

 

15) Display the list of all users in your database(use catalog table).

SQL>select username from all_users;

 

16) Display the names of all tables from current user;

SQL>select tname from tab;

 

17) Display the name of the current user.

SQL>show user;

 

18) Display the names of employees working in depart number 10 or 20 or 40 or employees working asCLERKS,SALESMAN or ANALYST.

SQL>select ename from emp where deptno in(10,20,40) or job in(‘CLERKS’,'SALESMAN’,'ANALYST’);

 

19) Display the names of employees whose name starts with alaphabet S.

SQL>select ename from emp where ename like ‘S%’;

 

20) Display the Employee names for employees whose name ends with alaphabet S.

SQL>select ename from emp where ename like ‘%S’;

 

21) Display the names of employees whose names have second alphabet A in their names.

SQL>select ename from emp where ename like ‘_A%’;

 

22) select the names of the employee whose names is exactly five characters in length.

SQL>select ename from emp where length(ename)=5;

 

23) Display the names of the employee who are not working as MANAGERS.

SQL>select ename from emp where job not in(‘MANAGER’);

 

24) Display the names of the employee who are not working as SALESMAN OR CLERK OR ANALYST.

SQL>select ename from emp where job not in(‘SALESMAN’,'CLERK’,'ANALYST’);

 

25) Display all rows from emp table.The system should wait after every screen full of informaction.

SQL>set pause on;

 

26) Display the total number of employee working in the company.

SQL>select count(*) from emp;

 

27) Display the total salary beiging paid to all employees.

SQL>select sum(sal) from emp;

 

28) Display the maximum salary from emp table.

SQL>select max(sal) from emp;

 

29) Display the minimum salary from emp table.

SQL>select min(sal) from emp;

 

30) Display the average salary from emp table.

SQL>select avg(sal) from emp;

 

31) Display the maximum salary being paid to CLERK.

SQL>select max(sal) from emp where job=’CLERK’;

 

32) Display the maximum salary being paid to depart number 20.

SQL>select max(sal) from emp where deptno=20;

 

33) Display the minimum salary being paid to any SALESMAN.

SQL>select min(sal) from emp where job=’SALESMAN’;

 

34) Display the average salary drawn by MANAGERS.

SQL>select avg(sal) from emp where job=’MANAGER’;

 

35) Display the total salary drawn by ANALYST working in depart number 40.

SQL>select sum(sal) from emp where job=’ANALYST’ and deptno=40;

 

36) Display the names of the employee in order of salary i.e the name of the employee earning lowest salary    should salary appear first.

SQL>select ename from emp order by sal;

 

37) Display the names of the employee in descending order of salary.

SQL>select ename from emp order by sal desc;

 

38) Display the names of the employee in order of employee name.

SQL>select ename from emp order by ename;

 

39) Display empno,ename,deptno,sal sort the output first base on name and within name by deptno and with in deptno by sal.

SQL>select empno,ename,deptno,sal from emp order by;

 

40) Display the name of the employee along with their annual salary(sal*12).The name of the employee earning highest annual salary should apper first.

SQL>select ename,sal*12 from emp order by sal desc;

 

41) Display name,salary,hra,pf,da,total salary for each employee. The output should be in the order of total salary,hra 15% of salary,da 10% of salary,pf 5% salary,total salary will be(salary+hra+da)-pf.

SQL>select ename,sal,sal/100*15 as hra,sal/100*5 as pf,sal/100*10 as da, sal+sal/100*15+sal/100*10-sal/100*5 as total from emp;

 

42) Display depart numbers and total number of employees working in each department.

SQL>select deptno,count(deptno)from emp group by deptno;

 

43) Display the various jobs and total number of employees within each job group.

SQL>select job,count(job)from emp group by job;

 

44) Display the depart numbers and total salary for each department.

SQL>select deptno,sum(sal) from emp group by deptno;

 

45) Display the depart numbers and max salary for each department.

SQL>select deptno,max(sal) from emp group by deptno;

 

46) Display the various jobs and total salary for each job

SQL>select job,sum(sal) from emp group by job;

 

47) Display the various jobs and total salary for each job

SQL>select job,min(sal) from emp group by job;

 

48) Display the depart numbers with more than three employees in each dept.

SQL>select deptno,count(deptno) from emp group by deptno having count(*)>3;

 

49) Display the various jobs along with total salary for each of the jobs where total salary is greater than 40000.

SQL>select job,sum(sal) from emp group by job having sum(sal)>40000;

 

50) Display the various jobs along with total number of employees in each job.The output should contain only those  jobs with more than three employees.

SQL>select job,count(empno) from emp group by job having count(job)>3;

 

51) Display the name of the empployee who earns highest salary.

SQL>select ename from emp where sal=(select max(sal) from emp);

 

52) Display the employee number and name for employee working as clerk and earning highest salary among clerks.

SQL>select empno,ename from emp where where job=’CLERK’ and sal=(select max(sal) from emp  where job=’CLERK’);

 

53) Display the names of salesman who earns a salary more than the highest salary of any clerk.

SQL>select ename,sal from emp where job=’SALESMAN’ and sal>(select max(sal) from emp where job=’CLERK’);

 

54) Display the names of clerks who earn a salary more than the lowest salary of any salesman.

SQL>select ename from emp where job=’CLERK’ and sal>(select min(sal) from emp where job=’SALESMAN’);

 

55) Display the names of the employees who earn highest salary in their respective departments.

SQL>select ename,sal,deptno from emp where sal in(select max(sal) from emp group by deptno);

 

56) Display the names of the employees who earn highest salaries in their respective job groups.

SQL>select ename,sal,job from emp where sal in(select max(sal) from emp group by job);

 

57) Display the employee names who are working in accounting department.

SQL>select ename from emp where deptno=(select deptno from dept wheredname=’ACCOUNTING’);

 

58) Display the employee names who are working in Ahmedabad .

SQL>select ename from emp where deptno=(select deptno from dept where LOC=’Ahmedabad’);

 

59) Display the Job groups having total salary greater than the maximum salary for managers.

SQL>SELECT JOB,SUM(SAL) FROM EMP GROUP BY JOB HAVING SUM(SAL)>(SELECT MAX(SAL) FROM EMP WHERE JOB=’MANAGER’);

 

60) Display the names of employees from department number 10 with salary grether than that of any employee working in other department.

SQL>select ename from emp where deptno=10 and sal>any(select sal from emp where deptno not in 10);

 

61) Display the names of the employees from department number 10 with salary greater than that of all employee working in other departments.

SQL>select ename from emp where deptno=10 and sal>all(select sal from emp where deptno not in 10);

 

62) Display the names of the employees in Uppercase.

SQL>select upper(ename)from emp;

 

63) Display the names of the employees in Lowecase.

SQL>select lower(ename)from emp;

 

64) Display the names of the employees in Propercase.

SQL>select initcap(ename)from emp;

 

65) Display the length of Your name using appropriate function.

SQL>select length(‘name’) from dual;

 

66) Display the length of all the employee names.

SQL>select length(ename) from emp;

 

67) select name of the employee concatenate with employee number.

SQL>select ename||empno from emp;

 

68) User appropriate function and extract 3 characters starting from 2 characters from the following  string ‘Oracle’. i.e the out put should be ‘ac’.

SQL>select substr(‘oracle’,3,2) from dual;

 

69) Find the First occurance of character ‘a’ from the following string i.e ‘Computer Maintenance Corporation’.

SQL>SELECT INSTR(‘Computer Maintenance Corporation’,'a’,1) FROM DUAL;

 

70) Replace every occurance of alphabhet A with B in the string Allens(use translate function)

SQL>select translate(‘Allens’,'A’,'B’) from dual;

 

71) Display the informaction from emp table.Where job manager is found it should be displayed as boos(Use replace function).

SQL>select replace(JOB,’MANAGER’,'BOSS’) FROM EMP;

 

72) Display empno,ename,deptno from emp table.Instead of display department numbers display the related department name(Use decode function).

SQL>select empno,ename,decode(deptno,10,’ACCOUNTING’,20,’RESEARCH’,30,’SALES’,40,’OPRATIONS’) from emp;

 

73) Display your age in days.

SQL>select to_date(sysdate)-to_date(’10-sep-77′)from dual;

 

74) Display your age in months.

SQL>select months_between(sysdate,’10-sep-77′) from dual;

 

75) Display the current date as 15th Augest Friday Nineteen Ninety Saven.

SQL>select to_char(sysdate,’ddth Month day year’) from dual;

 

76) Display the following output for each row from emp table. A has joined the company on wednesday 13th August ninten nintey.

SQL>select ENAME||’ HAS JOINED THE COMPANY ON  ‘||to_char(HIREDATE,’day ddth Month  year’)   from EMP;

 

77) Find the date for nearest saturday after current date.

SQL>SELECT NEXT_DAY(SYSDATE,’SATURDAY’)FROM DUAL;

 

78) Display current time.

SQL>select to_char(sysdate,’hh:MM:ss’) from dual.

79) Display the date three months Before the current date.

SQL>select add_months(sysdate,3) from dual;

 

80) Display the common jobs from department number 10 and 20.

SQL>select job from emp where deptno=10 and job in(select job from emp where deptno=20);

 

81)Display the names of employees who earn a salary more than that of A or that of salary grether than   that of B.

SQL>select ename,sal from emp where sal> (select sal from emp where ename=’A')and sal> (select sal from emp where ename=’B');

 

81) Display the jobs found in department 10 and 20 Eliminate duplicate jobs.

SQL>select distinct(job) from emp where deptno=10 or deptno=20;

(or)

SQL>select distinct(job) from emp where deptno in(10,20);

 

82) Display the jobs which are unique to department 10.

SQL>select distinct(job) from emp where deptno=10;

 

83) Display the details of those who do not have any person working under them.

SQL>select e.ename from emp,emp e where emp.mgr=e.empno group by e.ename havingcount(*)=1;

 

84) Display the details of those employees who are in sales department and grade is 3.

SQL>select * from emp where deptno=(select deptno from dept where dname=’SALES’)and sal between(select losal from salgrade where grade=3)and (select hisal from salgrade where grade=3);

 

85) Display those who are not managers and who are managers any one.

i)display the managers names

SQL>select distinct(m.ename) from emp e,emp m where m.empno=e.mgr;

ii)display the who are not managers

SQL>select ename from emp where ename not in(select distinct(m.ename) from emp e,emp m where m.empno=e.mgr);

 

86) Display those employee whose name contains not less than 4 characters.

SQL>select ename from emp where length(ename)>4;

 

87) Display those department whose name start with “S” while the location name ends with “K”.

SQL>select dname from dept where dname like ‘S%’ and loc like ‘%K’;

 

88) Display those employees whose manager name is akash.

SQL>select p.ename from emp e,emp p where e.empno=p.mgr and e.ename=’Akash’;

 

89) Display those employees whose salary is more than 3000 after giving 20% increment.

SQL>select ename,sal from emp where (sal+sal*.2)>3000;

 

90) Display all employees while their dept names;

SQL>select ename,dname from emp,dept where emp.deptno=dept.deptno;

 

91) Display ename who are working in sales dept.

SQL>select ename from emp where deptno=(select deptno from dept where dname=’SALES’);

 

92) Display employee name,deptname,salary and comm for those sal in between 2000 to 5000 while location is Ahmedabad .

SQL>select ename,dname,sal,comm from emp,dept where sal  between 2000 and 5000 and loc=’AHMEDABAD’ and emp.deptno=dept.deptno;

 

93)Display those employees whose salary greter than his manager salary.

SQL>select p.ename from emp e,emp p where e.empno=p.mgr and p.sal>e.sal;

 

94) Display those employees who are working in the same dept where his manager is work.

SQL>select p.ename from emp e,emp p where e.empno=p.mgr and p.deptno=e.deptno;

 

95) Display those employees who are not working under any manager.

SQL>select ename from emp where mgr is null;

 

96) Display grade and employees name for the dept no 10 or 30 but grade is not 4 while joined the company before 31-dec-82.

SQL>select ename,grade from emp,salgrade where sal between losal and hisal and deptno     in(10,30) and grade<>4 and hiredate<’31-DEC-82′;

 

97) Update the salary of each employee by 10% increment who are not eligiblw for commission.

SQL>update emp set sal=sal+sal*10/100 where comm is null;

 

98) SELECT those employee who joined the company before 31-dec-82 while their dept location is newyork or  Chicago.

SQL>SELECT EMPNO,ENAME,HIREDATE,DNAME,LOC FROM EMP,DEPT WHERE (EMP.DEPTNO=DEPT.DEPTNO)AND HIREDATE <’31-DEC-82′ AND DEPT.LOC IN(‘CHICAGO’,'NEW YORK’);

 

99) DISPLAY EMPLOYEE NAME,JOB,DEPARTMENT,LOCATION FOR ALL WHO ARE WORKING AS  MANAGER?

SQL>select ename,JOB,DNAME,LOCATION from emp,DEPT where mgr is not null;

 

100) DISPLAY THOSE EMPLOYEES WHOSE MANAGER NAME IS AKKI? –[AND ALSO DISPLAY THEIR MANAGER NAME]?

SQL> SELECT P.ENAME FROM EMP E, EMP P WHERE E.EMPNO=P.MGR AND E.ENAME=’AKKI’;

 

101) Display name and salary of ford if his salary is equal to hisal of his grade

SQL>select ename,sal,grade from emp,salgrade where sal between losal and hisal and ename =’FORD’ AND HISAL=SAL;

 

102) Display employee name,job,depart name ,manager name,his grade and make out an under department wise?

SQL>SELECT E.ENAME,E.JOB,DNAME,EMP.ENAME,GRADE FROM EMP,EMP E,SALGRADE,DEPT WHERE EMP.SAL BETWEEN LOSAL AND HISAL AND EMP.EMPNO=E.MGR AND EMP.DEPTNO=DEPT.DEPTNO ORDER BY DNAME;

 

103) List out all employees name,job,salary,grade and depart name for every one in the company  except ‘CLERK’.Sort on salary display the highest salary?

SQL>SELECT ENAME,JOB,DNAME,SAL,GRADE FROM EMP,SALGRADE,DEPT WHERE SAL BETWEEN LOSAL AND HISAL AND EMP.DEPTNO=DEPT.DEPTNO AND JOB NOT IN(‘CLERK’)ORDER BY SAL ASC;

 

104) Display the employee name,job and his manager.Display also employee who are without manager?

SQL>select e.ename,e.job,eMP.ename AS Manager from emp,emp e where emp.empno(+)=e.mgr;

 

105) Find out the top 5 earners of company?

SQL>SELECT DISTINCT SAL FROM EMP E WHERE 5>=(SELECT COUNT(DISTINCT SAL) FROM EMP A WHERE A.SAL>=E.SAL)ORDER BY SAL DESC;

 

106) Display name of those employee who are getting the highest salary?

SQL>select ename from emp where sal=(select max(sal) from emp);

 

107) Display those employee whose salary is equal to average of maximum and minimum?

SQL>select ename from emp where sal=(select max(sal)+min(sal)/2 from emp);

 

108) Select count of employee in each department  where count greater than 3?

SQL>select count(*) from emp group by deptno having count(deptno)>3;

 

109) Display dname where at least 3 are working and display only department name?

SQL>select distinct d.dname from dept d,emp e where d.deptno=e.deptno and 3>any (selectcount(deptno) from emp group by deptno);

 

110) Display name of those managers name whose salary is more than average salary of his company?

SQL>SELECT E.ENAME,EMP.ENAME FROM EMP,EMP E WHERE EMP.EMPNO=E.MGR AND E.SAL>(SELECT AVG(SAL) FROM EMP);

 

111)Display those managers name whose salary is more than average salary of his employee?

SQL>SELECT DISTINCT EMP.ENAME FROM EMP,EMP E WHERE

E.SAL <(SELECT AVG(EMP.SAL) FROM EMP

WHERE EMP.EMPNO=E.MGR GROUP BY EMP.ENAME) AND

EMP.EMPNO=E.MGR;

 

112) Display employee name,sal,comm and net pay for those employee whose net pay is greter than or equal to any other employee salary of the company?

SQL>select ename,sal,comm,sal+nvl(comm,0) as NetPay from emp where sal+nvl(comm,0) >any (select sal from emp);

 

113) Display all employees names with total sal of company with each employee name?

SQL>SELECT ENAME,(SELECT SUM(SAL)  FROM EMP) FROM EMP;

 

114) Find out last 5(least)earners of the company.?

SQL>SELECT DISTINCT SAL FROM EMP E WHERE

5>=(SELECT COUNT(DISTINCT SAL) FROM EMP A WHERE A.SAL<=E.SAL)

ORDER BY SAL DESC;

 

115) Find out the number of employees whose salary is greater than their manager salary?

SQL>SELECT E.ENAME FROM EMP ,EMP E WHERE EMP.EMPNO=E.MGR AND EMP.SAL<E.SAL;

 

116) Display those department where no employee working?

SQL>select dname from emp,dept where emp.deptno not in(emp.deptno);

 

117) Display those employee whose salary is ODD value?

SQL>select * from emp where sal<0;

 

118) Display those employee whose salary contains alleast 3 digits?

SQL>select * from emp where length(sal)>=3;

 

119) Display those employee who joined in the company in the month of Dec?

SQL>select ename from emp where to_char(hiredate,’MON’)=’DEC’;

 

120) Display those employees whose name contains “A”?

SQL>select ename from emp where instr(ename,’A')>0;

or

SQL>select ename from emp where ename like(‘%A%’);

 

121) Display those employee whose deptno is available in salary?

SQL>select emp.ename from emp, emp e where emp.sal=e.deptno;

 

122) Display those employee whose first 2 characters from hiredate -last 2 characters of salary?

SQL>select ename,SUBSTR(hiredate,1,2)||ENAME||substr(sal,-2,2) from emp;

 

123) Display those employee whose 10% of salary is equal to the year of joining?

SQL>select ename from emp where to_char(hiredate,’YY’)=sal*0.1;

 

124) Display those employee who are working in sales or research?

SQL>SELECT ENAME FROM EMP WHERE DEPTNO IN(SELECT DEPTNO FROM DEPT WHERE DNAME IN(‘SALES’,'RESEARCH’));

 

125) Display the grade of Akki?

SQL>SELECT ENAME,GRADE FROM EMP,SALGRADE WHERE SAL BETWEEN LOSAL AND HISAL AND Ename=’Akki’;

 

126) Display those employees who joined the company before 15 of the month?

SQL>select ename from emp where to_char(hiredate,’DD’)<15;

 

127) Display those employee who has joined before 15th of the month.

SQL>select ename from emp where to_char(hiredate,’DD’)<15;

 

128) Delete those records where no of employees in a particular department is less than 3.

SQL>delete from emp where deptno=(select deptno from emp group by deptno having count(deptno)<3);

 

129) Display the name of the department where no employee working.

SQL> SELECT E.ENAME,E.JOB,M.ENAME,M.JOB FROM EMP E,EMP M WHERE E.MGR=M.EMPNO;

 

130) Display those employees who are working as manager.

SQL>SELECT M.ENAME MANAGER FROM EMP M ,EMP E WHERE E.MGR=M.EMPNO GROUP BY M.ENAME;

 

131) Display those employees whose grade is equal to any number of sal but not equal to first number of sal?

SQL> SELECT ENAME,GRADE FROM EMP,SALGRADE WHERE GRADE NOT IN(SELECT SUBSTR(SAL,0,1)FROM EMP);

 

132) Print the details of all the employees who are Sub-ordinate to BLAKE?

SQL>select emp.ename from emp, emp e where emp.mgr=e.empno and e.ename=’BLAKE’;

 

133) Display employee name and his salary whose salary is greater than highest average of department number?

SQL>SELECT SAL FROM EMP WHERE SAL>(SELECT MAX(AVG(SAL)) FROM EMP GROUP BY DEPTNO);

 

134) Display the 10th record of emp table(without using rowid)

SQL>SELECT * FROM EMP WHERE ROWNUM<11 MINUS SELECT * FROM EMP WHERE ROWNUM<10;

 

135) Display the half of the ename’s in upper case and remaining lowercase?

SQL>SELECT SUBSTR(LOWER(ENAME),1,3)||SUBSTR(UPPER(ENAME),3,LENGTH(ENAME)) FROM EMP;

 

136) Display the 10th record of emp table without using group by and rowid?

SQL>SELECT * FROM EMP WHERE ROWNUM<11 MINUS SELECT * FROM EMP WHERE ROWNUM<10;

 

137) Create a copy of emp table;

SQL>create table new_table as select * from emp where 1=2;

 

138) Select ename if ename exists more than once.

SQL>select ename  from emp e group by ename having count(*)>1;

 

139) Display all enames in reverse order?(SMITH:HTIMS).

SQL>SELECT REVERSE(ENAME)FROM EMP;

 

140) Display those employee whose joining of month and grade is equal.

SQL>SELECT ENAME FROM EMP WHERE SAL BETWEEN

(SELECT LOSAL FROM SALGRADE WHERE

GRADE=TO_CHAR(HIREDATE,’MM’)) AND

(SELECT HISAL FROM SALGRADE WHERE

GRADE=TO_CHAR(HIREDATE,’MM’));

 

141) Display those employee whose joining DATE is available in deptno.

SQL>SELECT ENAME FROM EMP WHERE TO_CHAR(HIREDATE,’DD’)=DEPTNO;

 

142) Display those employees name as follows

A ALLEN

B BLAKE

SQL> SELECT SUBSTR(ENAME,1,1),ENAME FROM EMP;

 

143) List out the employees ename,sal,PF(20% OF SAL) from emp;

SQL>SELECT ENAME,SAL,SAL*.2 AS PF FROM EMP;

 

144) Create table emp with only one column empno;

SQL>Create table emp as select empno from emp where 1=2;

 

145) Add this column to emp table ename vrachar2(20).

SQL>alter table emp add(ename varchar2(20));

 

146) Oops I forgot give the primary key constraint.  Add in now.

SQL>alter table emp add primary key(empno);

 

147) Now increase the length of ename column to 30 characters.

SQL>alter table emp modify(ename varchar2(30));

 

148) Add salary column to emp table.

SQL>alter table emp add(sal number(10));

 

149) I want to give a validation saying that salary cannot be greater 10,000 (note give a name to this constraint)

SQL>alter table emp add constraint chk_001 check(sal<=10000);

 

150) For the time being I have decided that I will not impose this validation.My boss has agreed to pay more than 10,000.

SQL>again alter the table or drop constraint with  alter table emp drop constraint chk_001 (or)Disable the constraint by using  alter table emp modify constraint chk_001 disable;

 

151) My boss has changed his mind.  Now he doesn’t want to pay more than 10,000.so revoke that salary constraint.

SQL>alter table emp modify constraint chk_001 enable;

 

152) Add column called as mgr to your emp table;

SQL>alter table emp add(mgr number(5));

 

153) Oh! This column should be related to empno.  Give a command to add this constraint.

SQL>ALTER TABLE EMP ADD CONSTRAINT MGR_DEPT FOREIGN KEY(MGR) REFERENCESEMP(EMPNO);

 

154) Add deptno column to your emp table;

SQL>alter table emp add(deptno number(5));

 

155) This deptno column should be related to deptno column of dept table;

SQL>alter table emp add constraint dept_001 foreign key(deptno) reference dept(deptno) [deptno should be primary key];

 

156) Give the command to add the constraint.

SQL>alter table <table_name) add constraint <constraint_name> <constraint type>;

 

157) Create table called as newemp.  Using single command create this table as well as get data into this table(use create table as);

SQL>create table newemp as select * from emp;

SQL>Create table called as newemp.  This table should contain only

empno,ename,dname.

SQL>create table newemp as select empno,ename,dname from emp,dept where

1=2;

 

158) Delete the rows of employees who are working in the company for more than 2 years.

SQL>delete from emp where (sysdate-hiredate)/365>2;

 

159) Provide a commission(10% Comm Of Sal) to employees who are not earning any commission.

SQL>select sal*0.1 from emp where comm is null;

 

160) If any employee has commission his commission should be incremented by 10% of his salary.

SQL>update emp set comm=sal*.1 where comm is not null;

 

161) Display employee name and department name for each employee.

SQL>select empno,dname from emp,dept where emp.deptno=dept.deptno;

 

162)Display employee number,name and location of the department in which he is working.

SQL>select empno,ename,loc,dname from emp,dept where emp.deptno=dept.deptno;

 

163) Display ename,dname even if there are no employees working in a particular department(use outer join).

SQL>select ename,dname from emp,dept where emp.deptno=dept.deptno(+);

 

164) Display employee name and his manager name.

SQL>select p.ename,e.ename from emp e,emp p where e.empno=p.mgr;

 

165) Display the department name and total number of employees in each department.

SQL>select dname,count(ename) from emp,dept where emp.deptno=dept.deptno group by dname;

 

166)Display the department name along with total salary in each department.

SQL>select dname,sum(sal) from emp,dept where emp.deptno=dept.deptno group by dname;

 

167) Display itemname and total sales amount for each item.

SQL>select itemname,sum(amount) from item group by itemname;

 

168) Write a Query To Delete The Repeted Rows from emp table;

SQL>Delete from emp where rowid not in(select min(rowid)from emp group by ename);

 

169) TO DISPLAY 5 TO 7 ROWS FROM A TABLE

SQL>select ename from emp where rowid in(select rowid from emp where rownum<=7 minusselect rowid from empi where rownum<5);

 

170)  DISPLAY  TOP N ROWS FROM TABLE?

SQL>SELECT * FROM

(SELECT *  FROM EMP ORDER BY ENAME DESC)

WHERE ROWNUM <10;

 

171) DISPLAY   TOP 3 SALARIES FROM EMP;

SQL>SELECT SAL FROM ( SELECT  * FROM EMP ORDER  BY SAL DESC ) WHERE ROWNUM <4;

 

172) DISPLAY  9th FROM THE EMP TABLE?

SQL>SELECT ENAME FROM EMP

WHERE ROWID=(SELECT ROWID FROM EMP WHERE ROWNUM<=10

MINUS

SELECT ROWID FROM EMP WHERE ROWNUM <10);

 

173) select second max salary from emp

select max(sal) fromemp where sal<(select  max(sal) from emp);

 

174) Delete the 10th record of emp table.

SQL>DELETE FROM EMP WHERE EMPNO=(SELECT EMPNO FROM EMP WHERE ROWNUM<11 MINUS

SELECT EMPNO FROM EMP WHERE ROWNUM<10);

Oracle: 2 Understanding item_master table

posted Sep 3, 2011, 12:42 PM by Neil Mathew   [ updated Sep 3, 2011, 12:53 PM ]


ITEM_MASTER


ITNO NAME QOH CLA UOM ROL ROQ RATE
1090 Hammer 234 A pcs 12 34 400.9
1089 Saw 456 B pcs 17 23 800.89
1088 Lawn Mover 123 C pcs 21 21 5000.88
1087 Dish Washer 234 A pcs 76 45 950.87
1067 Baking Oven 145 A pcs 87 34 6000.67
1063 Spark Plug 150 C watt 34 67 750.63
1609 Alternator 168 B watt 50 56 750.09
1890 Battery 189 A volt 30 40 300.9
1378 Piston 234 B pcs 45 50 250.78

Qoh = Quantity on Hand
UoM = Unit of Measurement (pcs = pieces)
RoL = Reorder Level
RoQ = Reorder Quantity

 

Reorder Level

The Reorder Level represents the minimum stock level for the vehicle (equivalent to the MIN in a Min/Max-based system).  Whenever the vehicle drops at least 1 unit below the Reorder Level, auto-replenishment will automatically place an order with the central warehouse for shipment to the vehicle warehouse. (If the Reorder Level is set at 1, the technician will run out of that part before an order is placed. Therefore, it is recommended that you always set the Reorder Level to 2 or higher.)


A minimum amount of an item which a company holds in stock, such that, 
when stock falls to this amount, the item must be reordered


Example:

Warehouse

Part

Reorder Level

Reorder Quantity

FTSV092

12345678

2

3

Whenever the on-hand quantity of part 12345678 in vehicle FTSV092 drops below 2, a replenishment order will be created for parts to be shipped from the department’s central warehouse to this vehicle warehouse.


Reorder Quantity

The Reorder Quantity represents “how many more than the minimum are normally allowed” for this vehicle warehouse.

  • If you want to keep exactly a certain number on the vehicle, then the Reorder Quantity will be 0.
       For example, if you want to always keep 2 units of part XYZ on the vehicle, the reorder level is 2 and the reorder quantity is 0.


  • If you want to allow for some fluctuation in the on-hand quantity, then Reorder Quantity will be the difference between the minimum and maximum quantity to keep stocked.
       For example, if you want to always keep between 6 and 10 units of part XYZ on the vehicle, the reorder level is 6 and the reorder quantity is 4.


MAX stock level = Reorder Level + Reorder Quantity



Example:

Warehouse

Part

Reorder Level

Reorder Quantity

FTSV092

12345678

2

3


The maximum number to keep on hand [Reorder Level + Reorder Quantity] is 5. 


For part 12345678, the vehicle is to keep on-hand a quantity of between 2 and 5 at all times.  When placing an auto-replenishment order, the system will request the number needed to return the vehicle to a total of 5 for part 12345678.

If the vehicle reaches an on-hand quantity of 1, auto-replenishment will place an order for 4 more of this part.

 




Oracle: 1 Copy & Paste for Table Creation.

posted Sep 3, 2011, 9:08 AM by Neil Mathew   [ updated Oct 2, 2011, 11:49 PM ]

EMP


EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
7369 Smith Clerk 7902 17-DEC-80 800   20
7499 Allen Salesman 7698 20-FEB-81 1600 300 30
7521 Ward Salesman 7698 22-FEB-81 1250 500 30
7566 Jones Manager 7839 02-APR-81 2975   20
7654 Martin Salesman 7698 28-SEP-81 1250 1400 30
7698 Blake Manager 7839 01-MAY-81 2850   30
7782 Clark Manager 7839 09-JUN-81 2450   10
7788 Scott Analyst 7566 09-DEC-82 3000   20
7839 King President   17-NOV-81 5000   10
7844 Turner Salesman 7698 08-SEP-81 1500 0 30
7876 Adams Clerk 7788 03-DEC-89 950   30
7902 Ford Analyst 7566 04-DEC-81 3000   20

12 rows selected.


CREATE TABLE


CREATE TABLE EMP_nm ( 
 EMPNO NUMBER(4),
 ENAME VARCHAR2(20),
 JOB CHAR(10),
 MGR NUMBER(4),
 HIREDATE DATE,
 SAL NUMBER(9,2),
 COMM NUMBER(7,2),
 DEPTNO NUMBER(2)
 );


INSERT COMMAND:


INSERT INTO EMP_NM VALUES(7369,'Smith','Clerk',7902,'17-DEC-80',800,NULL,20);
INSERT INTO EMP_NM VALUES(7499,'Allen','Salesman',7698,'20-FEB-81',1600,300,30);
INSERT INTO EMP_NM VALUES(7521,'Ward','Salesman',7698,'22-FEB-81',1250,500,30);
INSERT INTO EMP_NM VALUES(7566,'Jones','Manager',7839,'02-APR-81',2975,NULL,20);
INSERT INTO EMP_NM VALUES(7654,'Martin','Salesman',7698,'28-SEP-81',1250,1400,30);
INSERT INTO EMP_NM VALUES(7698,'Blake','Manager',7839,'01-MAY-81',2850,NULL,30);
INSERT INTO EMP_NM VALUES(7782,'Clark','Manager',7839,'09-JUN-81',2450,NULL,10);
INSERT INTO EMP_NM VALUES(7788,'Scott','Analyst',7566,'09-DEC-82',3000,NULL,20);
INSERT INTO EMP_NM VALUES(7839,'King','President',NULL,'17-NOV-81',5000,NULL,10);
INSERT INTO EMP_NM VALUES(7844,'Turner','Salesman',7698,'08-SEP-81',1500,0,30);
INSERT INTO EMP_NM VALUES(7876,'Adams','Clerk',7788,'03-DEC-89',950,NULL,30);
INSERT INTO EMP_NM VALUES(7902,'Ford','Analyst',7566,'04-DEC-81',3000,NULL,20);



DEPT

DEPTNO DNAME LOC
10 Accounting New York
20 Research Dallas
30 Sales Chicago
40 Operations Boston

4 rows selected.


CREATE TABLE


CREATE TABLE DEPT_NM ( 
 DEPTNO NUMBER(2),
 DNAME varchar2(20),
 LOC varchar2(10)
 );


INSERT COMMAND:


INSERT INTO DEPT_NM VALUES(10,'Accounting','New York');
INSERT INTO DEPT_NM VALUES(20,'Research','Dallas');
INSERT INTO DEPT_NM VALUES(30,'Sales','Chicago');
INSERT INTO DEPT_NM VALUES(40,'Operations','Boston');



ITEM_MASTER

ITNO NAME QOH CLA UOM ROL ROQ RATE
1090 Hammer 234 A pcs 12 34 400.9
1089 Saw 456 B pcs 17 23 800.89
1088 Lawn Mover 123 C pcs 21 21 5000.88
1087 Dish Washer 234 A pcs 76 45 950.87
1067 Baking Oven 145 A pcs 87 34 6000.67
1063 Spark Plug 150 C watt 34 67 750.63
1609 Alternator 168 B watt 50 56 750.09
1890 Battery 189 A volt 30 40 300.9
1378 Piston 234 B pcs 45 50 250.78

9 rows selected.



CREATE TABLE


CREATE TABLE ITEM_MASTER (
ITNO NUMBER(4) PRIMARY KEY,
NAME VARCHAR2(20) NOT NULL,
QOH NUMBER(5) DEFAULT 100,
CLASS VARCHAR2(1) NOT NULL CHECK (CLASS IN ('A','B','C')),
UOM VARCHAR2(4),
ROL NUMBER(5),
ROQ NUMBER(5),
RATE NUMBER(8,2) NOT NULL
);

OR


CREATE TABLE ITEM_MASTER (
Itno number(4) primary key,
name varchar2(20) NOT NULL,
qoh number(5) DEFAULT 100,
class varchar2(1) NOT NULL,
uom varchar2(4),
rol number(5),
roq number(5),
rate number(8,2) NOT NULL,
CHECK (CLASS IN ('A','B','C'))
);



INSERT COMMAND:


INSERT INTO ITEM_MASTER VALUES (1090,'Hammer',234,'A','pcs',12,34,400.90);
INSERT INTO ITEM_MASTER VALUES (1089,'Saw',456,'B','pcs',17,23,800.89);
INSERT INTO ITEM_MASTER VALUES (1088,'Lawn Mover',123,'C','pcs',21,21,5000.88);
INSERT INTO ITEM_MASTER VALUES (1087,'Dish Washer',234,'A','pcs',76,45,950.87);
INSERT INTO ITEM_MASTER VALUES (1067,'Baking Oven',145,'A','pcs',87,34,6000.67);
INSERT INTO ITEM_MASTER VALUES (1063,'Spark Plug',150,'C','watt',34,67,750.63);
INSERT INTO ITEM_MASTER VALUES (1609,'Alternator',168,'B','watt',50,56,750.09);
INSERT INTO ITEM_MASTER VALUES (1890,'Battery',189,'A','volt',30,40,300.90);
INSERT INTO ITEM_MASTER VALUES (1378,'Piston',234,'B','pcs',45,50,250.78);




TRANSACTION


ITNO TYPE QTY RECEIPTNO DOT
1090 receive 500 A4333 01-JAN-09
1090 issue 100 A4336 23-FEB-10
1609 receive 215 A2143 23-FEB-10
1090 issue 150 A4343 12-MAR-10
1087 issue 300 B4143 01-SEP-10
1087 receive 50 A4143 16-DEC-10
1087 receive 50 A4144 20-DEC-10
1087 receive 50 A4145 30-DEC-10
1087 receive 50 A4149 11-JUN-11
1890 issue 25 C4143 15-JUL-11
1087 receive 50 A4151 20-JUL-11
1090 issue 200 A4133 01-AUG-11
1890 receive 15 C4113 10-SEP-11
1089 issue 125 C4041 19-SEP-11
1090 issue 120 A6336 25-SEP-11
1087 receive 50 A4155 26-SEP-11

16 rows selected.



CREATE TABLE:


CREATE TABLE TRANSACTION_NM (
ITNO NUMBER(4),
TYPE VARCHAR2(10),
QTY NUMBER(4),
RECEIPTNO VARCHAR2(20),
DOT DATE
);




INSERT COMMAND:


INSERT INTO TRANSACTION_NM VALUES(1090,'receive',500,'A4333','01-JAN-09');
INSERT INTO TRANSACTION_NM VALUES(1090,'issue',100,'A4336','23-FEB-10');
INSERT INTO TRANSACTION_NM VALUES(1609,'receive',215,'A2143','23-FEB-10');
INSERT INTO TRANSACTION_NM VALUES(1090,'issue',150,'A4343','12-MAR-10');
INSERT INTO TRANSACTION_NM VALUES(1087,'issue',300,'B4143','01-SEP-10');
INSERT INTO TRANSACTION_NM VALUES(1087,'receive',50,'A4143','16-DEC-10');
INSERT INTO TRANSACTION_NM VALUES(1087,'receive',50,'A4144','20-DEC-10');
INSERT INTO TRANSACTION_NM VALUES(1087,'receive',50,'A4145','30-DEC-10');
INSERT INTO TRANSACTION_NM VALUES(1087,'receive',50,'A4149','11-JUN-11');
INSERT INTO TRANSACTION_NM VALUES(1890,'issue',25,'C4143','15-JUL-11');
INSERT INTO TRANSACTION_NM VALUES(1087,'receive',50,'A4151','20-JUL-11');
INSERT INTO TRANSACTION_NM VALUES(1090,'issue',200,'A4133','01-AUG-11');
INSERT INTO TRANSACTION_NM VALUES(1890,'receive',15,'C4113','10-SEP-11');
INSERT INTO TRANSACTION_NM VALUES(1089,'issue',125,'C4041','19-SEP-11');
INSERT INTO TRANSACTION_NM VALUES(1090,'issue',120,'A6336','25-SEP-11');
INSERT INTO TRANSACTION_NM VALUES(1087,'receive',50,'A4155','26-SEP-11');



MySQL: 2 Creating Table

posted Aug 7, 2011, 3:43 AM by Neil Mathew   [ updated Aug 8, 2011, 9:38 AM ]



Once the database has been created, we can create the table. A shown below:

mysql> create database School
    -> ;
Query OK, 1 row affected (0.01 sec)

mysql> use school;
Database changed

mysql> create table emp
    -> ( Empno Integer,
    -> Ename varchar(20),
    -> Job char(10),
    -> Mgr Integer,
    -> Hiredate Date,
    -> Sal Decimal,
    -> Comm Decimal,
    -> DeptNo Integer
    -> );
Query OK, 0 rows affected (0.13 sec)

mysql> desc emp;
+----------+---------------+------+-----+---------+-------+
| Field    | Type          | Null | Key | Default | Extra |
+----------+---------------+------+-----+---------+-------+
| Empno    | int(11)       | YES  |     | NULL    |       |
| Ename    | varchar(20)   | YES  |     | NULL    |       |
| Job      | char(10)      | YES  |     | NULL    |       |
| Mgr      | int(11)       | YES  |     | NULL    |       |
| Hiredate | date          | YES  |     | NULL    |       |
| Sal      | decimal(10,0) | YES  |     | NULL    |       |
| Comm     | decimal(10,0) | YES  |     | NULL    |       |
| DeptNo   | int(11)       | YES  |     | NULL    |       |
+----------+---------------+------+-----+---------+-------+
8 rows in set (0.01 sec)
 
However, what concerned me was the difference in FORMATS. The Data Format in particular.
The MySQL date format is actually YYYY-MM-DD while
Oracle's is DD-MM-YYYY .

Also, unlike Oracle, the datatypes are different.
INT and DECIMAL would not be available in Oracle just as NUMBER is not available on MySQL.

Below is the table list copied from notepad. (MySQL format). The Tabs have been accordingly put.

7369	Smith	Clerk		7902	1980-12-17	800	NULL	20

7499	Allen	Salesman	7698	1981-02-20	1600	300	30					
7521	Ward	Salesman	7698	1981-02-22	1250	500	30					
7566	Jones	Manager		7839	1981-04-02	2975	NULL	20					
7654	Martin	Salesman	7698	1981-09-28	1250	1400	30					
7698	Blake	Manager		7839	1981-05-01	2850	NULL	30					
7782	Clark	Manager		7839	1981-06-09	2450	NULL	10					
7788	Scott	Analyst		7566	1982-12-09	3000	NULL	20					
7839	King	President	NULL	1981-11-17	5000	NULL	10					
7844	Turner	Salesman	7698	1981-09-08	1500	0	30					
7876	Adams	Clerk		7788	1989-12-03	950	NULL	30					
7902	Ford	Analyst		7566	1981-12-04	3000	NULL	20		

Manipulating the TABS to commas, and adding inverted commas to the String and Date entries, I've got one easy INSERT line.

mysql> INSERT INTO EMP
    -> VALUES (7499,'Allen','Salesman',7698,'1981-02-20',1600,300,30);
Query OK, 1 row affected (0.14 sec)

mysql> select * from emp;
+-------+-------+----------+------+------------+------+------+--------+
| Empno | Ename | Job      | Mgr  | Hiredate   | Sal  | Comm | DeptNo |
+-------+-------+----------+------+------------+------+------+--------+
|  7499 | Allen | Salesman | 7698 | 1981-02-20 | 1600 |  300 |     30 |
+-------+-------+----------+------+------------+------+------+--------+
1 row in set (0.05 sec)

Success. Now the rest:
 
mysql> INSERT INTO EMP VALUES(7521,'Ward','Salesman',7698,'1981-02-22',1250,500,
30);
Query OK, 1 row affected (0.01 sec)

mysql> INSERT INTO EMP VALUES(7566,'Jones','Manager',7839,'1981-04-02',2975,NULL
,20);
Query OK, 1 row affected (0.01 sec)

mysql> INSERT INTO EMP VALUES(7654,'Martin','Salesman',7698,'1981-09-28',1250,14
00,30);
Query OK, 1 row affected (0.01 sec)

mysql> INSERT INTO EMP VALUES(7698,'Blake','Manager',7839,'1981-05-01',2850,NULL
,30);
Query OK, 1 row affected (0.01 sec)

mysql> INSERT INTO EMP VALUES(7782,'Clark','Manager',7839,'1981-06-09',2450,NULL
,10);
Query OK, 1 row affected (0.01 sec)

mysql> INSERT INTO EMP VALUES(7788,'Scott','Analyst',7566,'1982-12-09',3000,NULL
,20);
Query OK, 1 row affected (0.01 sec)

mysql> INSERT INTO EMP VALUES(7839,'King','President',NULL,'1981-11-17',5000,NUL
L,10);
Query OK, 1 row affected (0.02 sec)

mysql> INSERT INTO EMP VALUES(7844,'Turner','Salesman',7698,'1981-09-08',1500,0,30);
Query OK, 1 row affected (0.01 sec)

mysql> INSERT INTO EMP VALUES(7876,'Adams','Clerk',7788,'1989-12-03',950,NULL,30
);
Query OK, 1 row affected (0.02 sec)

mysql> INSERT INTO EMP VALUES(7902,'Ford','Analyst',7566,'1981-12-04',3000,NULL,
20);
Query OK, 1 row affected (0.01 sec)

mysql> INSERT INTO EMP VALUES(7369,'Smith','Clerk',7902,'1980-12-17',800,NULL,20
);
Query OK, 1 row affected (0.01 sec)


mysql> select * from emp;
+-------+--------+-----------+------+------------+------+------+--------+
| EMPNO | ENAME  | JOB       | MGR  | HIREDATE   | SAL  | COMM | DEPTNO |
+-------+--------+-----------+------+------------+------+------+--------+
|  7499 | Allen  | Salesman  | 7698 | 1981-02-20 | 1600 |  300 |     30 |
|  7521 | Ward   | Salesman  | 7698 | 1981-02-22 | 1250 |  500 |     30 |
|  7566 | Jones  | Manager   | 7839 | 1981-04-02 | 2975 | NULL |     20 |
|  7654 | Martin | Salesman  | 7698 | 1981-09-28 | 1250 | 1400 |     30 |
|  7698 | Blake  | Manager   | 7839 | 1981-05-01 | 2850 | NULL |     30 |
|  7782 | Clark  | Manager   | 7839 | 1981-06-09 | 2450 | NULL |     10 |
|  7788 | Scott  | Analyst   | 7566 | 1982-12-09 | 3000 | NULL |     20 |
|  7839 | King   | President | NULL | 1981-11-17 | 5000 | NULL |     10 |
|  7844 | Turner | Salesman  | 7698 | 1981-09-08 | 1500 |    0 |     30 |
|  7876 | Adams  | Clerk     | 7788 | 1989-12-03 |  950 | NULL |     30 |
|  7902 | Ford   | Analyst   | 7566 | 1981-12-04 | 3000 | NULL |     20 |
|  7369 | Smith  | Clerk     | 7902 | 1980-12-17 |  800 | NULL |     20 |
+-------+--------+-----------+------+------------+------+------+--------+
12 rows in set (0.00 sec)


INSTANT COPY & PASTE FOR MYSQL USERS:


TABLE : EMP

Create TABLE code:

CREATE TABLE EMP ( 
 EMPNO INTEGER,
 ENAME VARCHAR(20),
 JOB CHAR(10),
 MGR INTEGER,
 HIREDATE DATE,
 SAL DECIMAL,
 COMM DECIMAL,
 DEPTNO INTEGER
 );

Values to insert:

INSERT INTO EMP VALUES(7369,'Smith','Clerk',7902,'1980-12-17',800,NULL,20);
INSERT INTO EMP VALUES(7499,'Allen','Salesman',7698,'1981-02-20',1600,300,30);
INSERT INTO EMP VALUES(7521,'Ward','Salesman',7698,'1981-02-22',1250,500,30);
INSERT INTO EMP VALUES(7566,'Jones','Manager',7839,'1981-04-02',2975,NULL,20);
INSERT INTO EMP VALUES(7654,'Martin','Salesman',7698,'1981-09-28',1250,1400,30);
INSERT INTO EMP VALUES(7698,'Blake','Manager',7839,'1981-05-01',2850,NULL,30);
INSERT INTO EMP VALUES(7782,'Clark','Manager',7839,'1981-06-09',2450,NULL,10);
INSERT INTO EMP VALUES(7788,'Scott','Analyst',7566,'1982-12-09',3000,NULL,20);
INSERT INTO EMP VALUES(7839,'King','President',NULL,'1981-11-17',5000,NULL,10);
INSERT INTO EMP VALUES(7844,'Turner','Salesman',7698,'1981-09-08',1500,0,30);
INSERT INTO EMP VALUES(7876,'Adams','Clerk',7788,'1989-12-03',950,NULL,30);
INSERT INTO EMP VALUES(7902,'Ford','Analyst',7566,'1981-12-04',3000,NULL,20);



TABLE : DEPT

Create TABLE code:

CREATE TABLE DEPT ( 
 DEPTNO INTEGER,
 DNAME varchar(20),
 LOC varchar(20)
 );

Values to insert:

INSERT INTO DEPT VALUES(10,'Accounting','New York');

INSERT INTO DEPT VALUES(20,'Research','Dallas');

INSERT INTO DEPT VALUES(30,'Sales','Chicago');

INSERT INTO DEPT VALUES(40,'Operations','Boston');


Common : Varchar, Varchar2, Char

posted Aug 7, 2011, 12:55 AM by Neil Mathew   [ updated Aug 7, 2011, 1:02 AM ]

Basic Difference between the three:

CHAR(5) is fixed length, right padded with spaces.

VARCHAR(5) is fixed length, right padded with null

VARCHAR2(5) is variable length. (Oracle)

( But, remember CHAR is faster than VARCHAR - some times up to 50% faster. )

Difference between Varchar & Varchar2:

(1) 
Varchar can have MAximum 2000 character while
Varchar can contain maximum 4000 character.

(2)
Varchar is of ANSI SQL standard.
Varchar2 is of Oracle standard.

(3) 
Varchar2 Datatype: The string value's length will be stored on disk with the value itself.

VARCHAR will occupy space for NULL values
VARCHAR2 datatype will not occupy any space.



Oracle v/s MySQL: Datatypes

posted Aug 7, 2011, 12:33 AM by Neil Mathew   [ updated Nov 10, 2011, 12:56 AM ]

Comparing the Datatypes of the two. (Click for Source of Info)



NUMERIC


MySQLSizeOracle

TINYINT

1 Byte

NUMBER(3,0)

SMALLINT

2 Bytes

NUMBER(5,0)

MEDIUMINT

3 Bytes

NUMBER (7,0)`

INT

4 Bytes

NUMBER (10,0)

INTEGER

4 Bytes

NUMBER (10,0)

BIGINT

8 Bytes

NUMBER (19,0)

FLOAT(X<=24)

4 Bytes

FLOAT(0)

FLOAT(25<=X <=53)

8 Bytes

FLOAT(24)

DOUBLE

8 Bytes

FLOAT(24)

DOUBLE PRECION

8 Bytes

FLOAT(24)

REAL

8 Bytes

FLOAT(24)

DECIMAL

M Bytes(D+2, if M<D)

FLOAT(24)

NUMERIC

M Bytes(D+2, if M<D)

NUMBER



Note: Noticed in oracle:

NUMBER ( 5, 2 ) is a floating type number which allows a column to have a 5digit number with two decimal places. That is, only 3 non-decimal places are allowed. eg: 300.02

I had some errors arising because I entered less digits in the decimal places or too many overall. Just a note.

STRING


MySQLSizeOracle

CHAR(m)

M Bytes, 1<=M<=255

CHAR

VARCHAR(m)

L+1 Bytes whereas L<=M and 1<=M<=255

VARCHAR2

TINYBLOB

L + 1 Bytes whereas L<2 ^8

RAW, BLOB

BLOB

L + 2 Bytes whereas L<2^16

RAW, BLOB

TEXT

L + 2 Bytes whereas L<2^16

RAW, BLOB

MEDIUMBLOB

L + 3 Bytes whereas L < 2^ 24

RAW, BLOB

MEDIUMTEXT

L + 3 Bytes whereas L < 2^ 24

RAW, BLOB

LONGBLOB

L + 4 Bytes whereas L < 2 ^ 32

RAW, BLOB

LONGTEXT

L + 4 Bytes whereas L < 2 ^ 32

RAW, BLOB



DATE & TIME:


MySQLSizeOracle

DATE

3 Bytes

DATE

DATETIME

8 Bytes

DATE

TIMESTAMP

4 Bytes

NUMBER

TIME

3 Bytes

DATE

YEAR

1 Byte

NUMBER



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