Oracle PL/SQL Tutorial/Analytical Functions/GROUPING ID
An Application of GROUPING_ID()
- GROUPING_ID() function filters rows using a HAVING clause to exclude rows that don"t contain a subtotal or total.
- GROUPING_ID() function accepts one or more columns and returns the decimal equivalent of the GROUPING bit vector.
- GROUPING bit vector is computed by combining the results of a call to the GROUPING() function for each column in order.
- One application of GROUPING_ID() is to filter rows using a HAVING clause.
- Your HAVING clause can exclude rows that don"t contain a subtotal or total by simply checking if GROUPING_ID() returns a value greater than 0.
Quote from:
Oracle Database 10g SQL (Osborne ORACLE Press Series) (Paperback)
# Paperback: 608 pages
# Publisher: McGraw-Hill Osborne Media; 1st edition (February 20, 2004)
# Language: English
# ISBN-10: 0072229810
# ISBN-13: 978-0072229813
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>
SQL>
SQL> SELECT
2 city, description,
3 GROUPING_ID(city, description) AS grp_id,
4 SUM(salary)
5 FROM employee
6 GROUP BY CUBE(city, description)
7 HAVING GROUPING_ID(city, description) > 0;
CITY DESCRIPTION GRP_ID SUM(SALARY)
---------- --------------- ---------- -----------
3 32574.02
Tester 2 21096.9
Manager 2 10242.56
Programmer 2 1234.56
Toronto 1 1234.56
New York 1 12220.56
Vancouver 1 19118.9
7 rows selected.
SQL>
SQL>
SQL> -- clean the table
SQL> drop table Employee
2 /
Table dropped.
SQL>
SQL>
group_id with cube
SQL> create table employees(
2 empno NUMBER(4)
3 , ename VARCHAR2(8)
4 , init VARCHAR2(5)
5 , job VARCHAR2(8)
6 , mgr NUMBER(4)
7 , bdate DATE
8 , msal NUMBER(6,2)
9 , comm NUMBER(6,2)
10 , deptno NUMBER(2) ) ;
Table created.
SQL>
SQL>
SQL> insert into employees values(1,"Jason", "N", "TRAINER", 2, date "1965-12-18", 800 , NULL, 10);
1 row created.
SQL> insert into employees values(2,"Jerry", "J", "SALESREP",3, date "1966-11-19", 1600, 300, 10);
1 row created.
SQL> insert into employees values(3,"Jord", "T" , "SALESREP",4, date "1967-10-21", 1700, 500, 20);
1 row created.
SQL> insert into employees values(4,"Mary", "J", "MANAGER", 5, date "1968-09-22", 1800, NULL, 20);
1 row created.
SQL> insert into employees values(5,"Joe", "P", "SALESREP",6, date "1969-08-23", 1900, 1400, 30);
1 row created.
SQL> insert into employees values(6,"Black", "R", "MANAGER", 7, date "1970-07-24", 2000, NULL, 30);
1 row created.
SQL> insert into employees values(7,"Red", "A", "MANAGER", 8, date "1971-06-25", 2100, NULL, 40);
1 row created.
SQL> insert into employees values(8,"White", "S", "TRAINER", 9, date "1972-05-26", 2200, NULL, 40);
1 row created.
SQL> insert into employees values(9,"Yellow", "C", "DIRECTOR",10, date "1973-04-27", 2300, NULL, 20);
1 row created.
SQL> insert into employees values(10,"Pink", "J", "SALESREP",null,date "1974-03-28", NULL, 0, 30);
1 row created.
SQL>
SQL>
SQL> select deptno, job
2 , grouping_id(deptno, job) gid
3 from employees
4 group by CUBE(deptno, job);
DEPTNO JOB GID
---------- -------- ----------
3
MANAGER 2
TRAINER 2
DIRECTOR 2
SALESREP 2
10 1
10 TRAINER 0
10 SALESREP 0
20 1
20 MANAGER 0
20 DIRECTOR 0
20 SALESREP 0
30 1
30 MANAGER 0
30 SALESREP 0
40 1
40 MANAGER 0
40 TRAINER 0
18 rows selected.
SQL>
SQL> drop table employees;
Table dropped.
grouping_id function with case statement
SQL>
SQL> create table employees(
2 empno NUMBER(4)
3 , ename VARCHAR2(8)
4 , init VARCHAR2(5)
5 , job VARCHAR2(8)
6 , mgr NUMBER(4)
7 , bdate DATE
8 , msal NUMBER(6,2)
9 , comm NUMBER(6,2)
10 , deptno NUMBER(2) ) ;
Table created.
SQL>
SQL>
SQL> insert into employees values(1,"Jason", "N", "TRAINER", 2, date "1965-12-18", 800 , NULL, 10);
1 row created.
SQL> insert into employees values(2,"Jerry", "J", "SALESREP",3, date "1966-11-19", 1600, 300, 10);
1 row created.
SQL> insert into employees values(3,"Jord", "T" , "SALESREP",4, date "1967-10-21", 1700, 500, 20);
1 row created.
SQL> insert into employees values(4,"Mary", "J", "MANAGER", 5, date "1968-09-22", 1800, NULL, 20);
1 row created.
SQL> insert into employees values(5,"Joe", "P", "SALESREP",6, date "1969-08-23", 1900, 1400, 30);
1 row created.
SQL> insert into employees values(6,"Black", "R", "MANAGER", 7, date "1970-07-24", 2000, NULL, 30);
1 row created.
SQL> insert into employees values(7,"Red", "A", "MANAGER", 8, date "1971-06-25", 2100, NULL, 40);
1 row created.
SQL> insert into employees values(8,"White", "S", "TRAINER", 9, date "1972-05-26", 2200, NULL, 40);
1 row created.
SQL> insert into employees values(9,"Yellow", "C", "DIRECTOR",10, date "1973-04-27", 2300, NULL, 20);
1 row created.
SQL> insert into employees values(10,"Pink", "J", "SALESREP",null,date "1974-03-28", NULL, 0, 30);
1 row created.
SQL>
SQL>
SQL> select deptno
2 , case grouping_id(deptno,job)
3 when 0 then job
4 when 1 then "**dept **"
5 when 3 then "**total**"
6 end job
7 , count(empno) headcount
8 from employees
9 group by ROLLUP(deptno, job);
DEPTNO JOB HEADCOUNT
---------- --------- ----------
10 TRAINER 1
10 SALESREP 1
10 **dept ** 2
20 MANAGER 1
20 DIRECTOR 1
20 SALESREP 1
20 **dept ** 3
30 MANAGER 1
30 SALESREP 2
30 **dept ** 3
40 MANAGER 1
40 TRAINER 1
40 **dept ** 2
**total** 10
14 rows selected.
SQL>
SQL> drop table employees;
Table dropped.
SQL>