GROUP_ID() function can remove duplicate rows returned by a GROUP BY clause.
GROUP_ID() doesn't accept any parameters.
If n duplicates exist for a particular grouping, GROUP_ID returns numbers in the range 0 to n-1.
SQL> -- create demo table
SQL> create table Employee(
2 ID VARCHAR2(4 BYTE) NOT NULL,
3 First_Name VARCHAR2(10 BYTE),
4 Last_Name VARCHAR2(10 BYTE),
5 Start_Date DATE,
6 End_Date DATE,
7 Salary Number(8,2),
8 City VARCHAR2(10 BYTE),
9 Description VARCHAR2(15 BYTE)
10 )
11 /
Table created.
SQL>
SQL> -- prepare data
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values ('01','Jason', 'Martin', to_date('19960725','YYYYMMDD'), to_date('20060725','YYYYMMDD'), 1234.56, 'Toronto', 'Programmer')
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values('02','Alison', 'Mathews', to_date('19760321','YYYYMMDD'), to_date('19860221','YYYYMMDD'), 6661.78, 'Vancouver','Tester')
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values('03','James', 'Smith', to_date('19781212','YYYYMMDD'), to_date('19900315','YYYYMMDD'), 6544.78, 'Vancouver','Tester')
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values('04','Celia', 'Rice', to_date('19821024','YYYYMMDD'), to_date('19990421','YYYYMMDD'), 2344.78, 'Vancouver','Manager')
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values('05','Robert', 'Black', to_date('19840115','YYYYMMDD'), to_date('19980808','YYYYMMDD'), 2334.78, 'Vancouver','Tester')
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values('06','Linda', 'Green', to_date('19870730','YYYYMMDD'), to_date('19960104','YYYYMMDD'), 4322.78,'New York', 'Tester')
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values('07','David', 'Larry', to_date('19901231','YYYYMMDD'), to_date('19980212','YYYYMMDD'), 7897.78,'New York', 'Manager')
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values('08','James', 'Cat', to_date('19960917','YYYYMMDD'), to_date('20020415','YYYYMMDD'), 1232.78,'Vancouver', 'Tester')
3 /
1 row created.
SQL>
SQL>
SQL>
SQL> -- display data in the table
SQL> select * from Employee
2 /
ID FIRST_NAME LAST_NAME START_DAT END_DATE SALARY CITY DESCRIPTION
---- ---------- ---------- --------- --------- ---------- ---------- ---------------
01 Jason Martin 25-JUL-96 25-JUL-06 1234.56 Toronto Programmer
02 Alison Mathews 21-MAR-76 21-FEB-86 6661.78 Vancouver Tester
03 James Smith 12-DEC-78 15-MAR-90 6544.78 Vancouver Tester
04 Celia Rice 24-OCT-82 21-APR-99 2344.78 Vancouver Manager
05 Robert Black 15-JAN-84 08-AUG-98 2334.78 Vancouver Tester
06 Linda Green 30-JUL-87 04-JAN-96 4322.78 New York Tester
07 David Larry 31-DEC-90 12-FEB-98 7897.78 New York Manager
08 James Cat 17-SEP-96 15-APR-02 1232.78 Vancouver Tester
8 rows selected.
SQL>
SQL> SELECT city, description, GROUP_ID(), SUM(salary)
2 FROM employee
3 GROUP BY city, ROLLUP(city, description);
CITY DESCRIPTION GROUP_ID() SUM(SALARY)
---------- --------------- ---------- -----------
Toronto Programmer 0 1234.56
New York Tester 0 4322.78
New York Manager 0 7897.78
Vancouver Tester 0 16774.12
Vancouver Manager 0 2344.78
Toronto 0 1234.56
New York 0 12220.56
Vancouver 0 19118.9
Toronto 1 1234.56
New York 1 12220.56
Vancouver 1 19118.9
11 rows selected.
SQL>
SQL>
SQL>
SQL> -- clean the table
SQL> drop table Employee
2 /
Table dropped.
SQL>
SQL>
SQL>