GROUPING() returns 0 for the rows that have non-null values, and 1 for the last row that has a null value : GROUPING « Analytical Functions « Oracle PL / SQL






GROUPING() returns 0 for the rows that have non-null values, and 1 for the last row that has a null value

 

SQL>
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_DATE  END_DATE        SALARY CITY       DESCRIPTION
---- ---------- ---------- ----------- ----------- ---------- ---------- ---------------
01   Jason      Martin     1996-JUL-25 2006-JUL-25    1234.56 Toronto    Programmer
02   Alison     Mathews    1976-MAR-21 1986-FEB-21    6661.78 Vancouver  Tester
03   James      Smith      1978-DEC-12 1990-MAR-15    6544.78 Vancouver  Tester
04   Celia      Rice       1982-OCT-24 1999-APR-21    2344.78 Vancouver  Manager
05   Robert     Black      1984-JAN-15 1998-AUG-08    2334.78 Vancouver  Tester
06   Linda      Green      1987-JUL-30 1996-JAN-04    4322.78 New York   Tester
07   David      Larry      1990-DEC-31 1998-FEB-12    7897.78 New York   Manager
08   James      Cat        1996-SEP-17 2002-APR-15    1232.78 Vancouver  Tester

8 rows selected.

SQL>
SQL>
SQL> --GROUPING() returns 0 for the rows that have non-null values, and 1 for the last row that has a null value:
SQL>
SQL> SELECT GROUPING(city), city, SUM(salary)
  2  FROM employee
  3  GROUP BY ROLLUP(city);

GROUPING(CITY) CITY       SUM(SALARY)
-------------- ---------- -----------
             0 New York      12220.56
             0 Toronto        1234.56
             0 Vancouver      19118.9
             1               32574.02

SQL>
SQL>
SQL>
SQL> -- clean the table
SQL> drop table Employee
  2  /

Table dropped.

SQL>
SQL>
SQL>

           
         
  








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