SQL Server/T-SQL/Aggregate Functions/Aggregate Function Basics

Null value is eliminated by an aggregate or other SET operation

1> create table employee(
2>     ID          int,
3>     name        nvarchar (10),
4>     salary      int,
5>     start_date  datetime,
6>     city        nvarchar (10),
7>     region      char (1))
8> GO
1>
2> insert into employee (ID, name,    salary, start_date, city,       region)
3>               values (1,  "Jason", 40420,  "02/01/94", "New York", "W")
4> GO
(1 rows affected)
1> insert into employee (ID, name,    salary, start_date, city,       region)
2>               values (2,  "Robert",14420,  "01/02/95", "Vancouver","N")
3> GO
(1 rows affected)
1> insert into employee (ID, name,    salary, start_date, city,       region)
2>               values (3,  "Celia", 24020,  "12/03/96", "Toronto",  "W")
3> GO
(1 rows affected)
1> insert into employee (ID, name,    salary, start_date, city,       region)
2>               values (4,  "Linda", 40620,  "11/04/97", "New York", "N")
3> GO
(1 rows affected)
1> insert into employee (ID, name,    salary, start_date, city,       region)
2>               values (5,  "David", 80026,  "10/05/98", "Vancouver","W")
3> GO
(1 rows affected)
1> insert into employee (ID, name,    salary, start_date, city,       region)
2>               values (6,  "James", 70060,  "09/06/99", "Toronto",  "N")
3> GO
(1 rows affected)
1> insert into employee (ID, name,    salary, start_date, city,       region)
2>               values (7,  "Alison",90620,  "08/07/00", "New York", "W")
3> GO
(1 rows affected)
1> insert into employee (ID, name,    salary, start_date, city,       region)
2>               values (8,  "Chris", 26020,  "07/08/01", "Vancouver","N")
3> GO
(1 rows affected)
1> insert into employee (ID, name,    salary, start_date, city,       region)
2>               values (9,  "Mary",  60020,  "06/09/02", "Toronto",  "W")
3> GO
(1 rows affected)
1>
2> select * from employee
3> GO
ID          name       salary      start_date              city       region
----------- ---------- ----------- ----------------------- ---------- ------
          1 Jason            40420 1994-02-01 00:00:00.000 New York   W
          2 Robert           14420 1995-01-02 00:00:00.000 Vancouver  N
          3 Celia            24020 1996-12-03 00:00:00.000 Toronto    W
          4 Linda            40620 1997-11-04 00:00:00.000 New York   N
          5 David            80026 1998-10-05 00:00:00.000 Vancouver  W
          6 James            70060 1999-09-06 00:00:00.000 Toronto    N
          7 Alison           90620 2000-08-07 00:00:00.000 New York   W
          8 Chris            26020 2001-07-08 00:00:00.000 Vancouver  N
          9 Mary             60020 2002-06-09 00:00:00.000 Toronto    W
(9 rows affected)
1>
2> SELECT COUNT(Name)
3> FROM Employee
4> GO
-----------
          9
(1 rows affected)
1>
2> update Employee set name = null;
3> GO
(9 rows affected)
1>
2> SELECT COUNT(Name)
3> FROM Employee
4> GO
-----------
          0
Warning: Null value is eliminated by an aggregate or other SET operation.
1>
2>
3>
4>
5>
6> drop table employee
7> GO
1>

Use sum function in a table join

1> create table employee(
2>     ID          int,
3>     name        nvarchar (10),
4>     salary      int )
5> GO
1>
2> create table job(
3>     ID              int,
4>     title nvarchar  (10),
5>     averageSalary   int)
6> GO
1>
2>
3> insert into employee (ID, name, salary) values (1,  "Jason", 1234)
4> GO
1> insert into employee (ID, name, salary) values (2,  "Robert", 4321)
2> GO
1> insert into employee (ID, name, salary) values (3,  "Celia", 5432)
2> GO
1> insert into employee (ID, name, salary) values (4,  "Linda", 3456)
2> GO
1> insert into employee (ID, name, salary) values (5,  "David", 7654)
2> GO
1> insert into employee (ID, name, salary) values (6,  "James", 4567)
2> GO
1> insert into employee (ID, name, salary) values (7,  "Alison", 8744)
2> GO
1> insert into employee (ID, name, salary) values (8,  "Chris", 9875)
2> GO
1> insert into employee (ID, name, salary) values (9,  "Mary", 2345)
2> GO
1>
2> insert into job(ID, title, averageSalary) values(1,"Developer",3000)
3> GO
1> insert into job(ID, title, averageSalary) values(2,"Tester", 4000)
2> GO
1> insert into job(ID, title, averageSalary) values(3,"Designer", 5000)
2> GO
1> insert into job(ID, title, averageSalary) values(4,"Programmer", 6000)
2> GO
1>
2>
3> select * from employee;
4> GO
ID          name       salary
----------- ---------- -----------
          1 Jason             1234
          2 Robert            4321
          3 Celia             5432
          4 Linda             3456
          5 David             7654
          6 James             4567
          7 Alison            8744
          8 Chris             9875
          9 Mary              2345
1> select * from job;
2> GO
ID          title      averageSalary
----------- ---------- -------------
          1 Developer           3000
          2 Tester              4000
          3 Designer            5000
          4 Programmer          6000
1>
2> SELECT sum(e.salary), j.title
3> FROM Employee e CROSS JOIN job j
4> group by j.title
5> GO
            title
----------- ----------
      47628 Designer
      47628 Developer
      47628 Programmer
      47628 Tester
1>
2>
3> drop table employee;
4> drop table job;
5> GO
1>
2>