GROUPING() Comes to the Rescue : GROUPING « Analytical Functions « SQL Server / T-SQL Tutorial






You use the GROUPING() function with an expression that might contain NULLs.
GROUPING() function returning value is only relevant when the value is NULL.
GROUPING() function returns 1 when the NULL represents a super aggregate.
GROUPING() function returns 0 when the NULL represents a group of NULLs in the base table.

CUBE Query and the GROUPING() function
10>
11> CREATE TABLE employee(
12>    id          INTEGER NOT NULL PRIMARY KEY,
13>    first_name  VARCHAR(10),
14>    last_name   VARCHAR(10),
15>    salary      DECIMAL(10,2),
16>    start_Date  DATETIME,
17>    region      VARCHAR(10),
18>    city        VARCHAR(20),
19>    managerid   INTEGER
20> );
21> GO
1> INSERT INTO employee VALUES (1, 'Jason' ,  'Martin', 5890,'2005-03-22','North','Vancouver',3);
2> GO

(1 rows affected)
1> INSERT INTO employee VALUES (2, 'Alison',  'Mathews',4789,'2003-07-21','South','Utown',4);
2> GO

(1 rows affected)
1> INSERT INTO employee VALUES (3, 'James' ,  'Smith',  6678,'2001-12-01','North','Paris',5);
2> GO

(1 rows affected)
1> INSERT INTO employee VALUES (4, 'Celia' ,  'Rice',   5567,'2006-03-03','South','London',6);
2> GO

(1 rows affected)
1> INSERT INTO employee VALUES (5, 'Robert',  'Black',  4467,'2004-07-02','East','Newton',7);
2> GO

(1 rows affected)
1> INSERT INTO employee VALUES (6, 'Linda' ,  'Green' , 6456,'2002-05-19','East','Calgary',8);
2> GO

(1 rows affected)
1> INSERT INTO employee VALUES (7, 'David' ,  'Larry',  5345,'2008-03-18','West','New York',9);
2> GO

(1 rows affected)
1> INSERT INTO employee VALUES (8, 'James' ,  'Cat',    4234,'2007-07-17','West','Regina',9);
2> GO

(1 rows affected)
1> INSERT INTO employee VALUES (9, 'Joan'  ,  'Act',    6123,'2001-04-16','North','Toronto',10);
2> GO

(1 rows affected)
1>
2> select * from employee;
3> GO
id          first_name last_name  salary       start_Date              region     city                 managerid
----------- ---------- ---------- ------------ ----------------------- ---------- -------------------- -----------
          1 Jason      Martin          5890.00 2005-03-22 00:00:00.000 North      Vancouver                      3
          2 Alison     Mathews         4789.00 2003-07-21 00:00:00.000 South      Utown                          4
          3 James      Smith           6678.00 2001-12-01 00:00:00.000 North      Paris                          5
          4 Celia      Rice            5567.00 2006-03-03 00:00:00.000 South      London                         6
          5 Robert     Black           4467.00 2004-07-02 00:00:00.000 East       Newton                         7
          6 Linda      Green           6456.00 2002-05-19 00:00:00.000 East       Calgary                        8
          7 David      Larry           5345.00 2008-03-18 00:00:00.000 West       New York                       9
          8 James      Cat             4234.00 2007-07-17 00:00:00.000 West       Regina                         9
          9 Joan       Act             6123.00 2001-04-16 00:00:00.000 North      Toronto                       10

(9 rows affected)
1>
2>
3>
4> SELECT
5>   ID,
6>   GROUPING(ID)      AS Grp_Cust,
7>   YEAR(Start_Date)           AS Order_Year,
8>   GROUPING(YEAR(Start_Date)) AS Grp_Year,
9>   COUNT(*) as Order_Count
10> FROM
11>   Employee
12> GROUP BY
13>   ID,
14>   YEAR(Start_Date)
15> WITH CUBE
16> GO
ID          Grp_Cust Order_Year  Grp_Year Order_Count
----------- -------- ----------- -------- -----------
          1        0        2005        0           1
          1        0        NULL        1           1
          2        0        2003        0           1
          2        0        NULL        1           1
          3        0        2001        0           1
          3        0        NULL        1           1
          4        0        2006        0           1
          4        0        NULL        1           1
          5        0        2004        0           1
          5        0        NULL        1           1
          6        0        2002        0           1
          6        0        NULL        1           1
          7        0        2008        0           1
          7        0        NULL        1           1
          8        0        2007        0           1
          8        0        NULL        1           1
          9        0        2001        0           1
          9        0        NULL        1           1
       NULL        1        NULL        1           9
       NULL        1        2001        0           2
       NULL        1        2002        0           1
       NULL        1        2003        0           1
       NULL        1        2004        0           1
       NULL        1        2005        0           1
       NULL        1        2006        0           1
       NULL        1        2007        0           1
       NULL        1        2008        0           1

(27 rows affected)
1>
2> drop table employee;
3> GO








14.4.GROUPING
14.4.1.GROUPING() Comes to the Rescue
14.4.2.Replacing NULLs with ALL and UNKNOWN
14.4.3.The syntax of the GROUPING function: GROUPING(column_name)
14.4.4.GROUPING() function returns a bit value (1 or 0) to indicate that a row is a rollup.