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>