SQL Server/T-SQL/Transact SQL/Select Data — различия между версиями
Admin (обсуждение | вклад) м (1 версия) |
Admin (обсуждение | вклад) м (1 версия) |
(нет различий)
|
Текущая версия на 10:20, 26 мая 2010
Do calculation in a procedure
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> Drop procedure spMarkupTest
3> GO
1>
2> CREATE PROC spMarkupTest
3> @MarkupAsPercent money
4> AS
5> DECLARE @Multiplier money
6> SELECT @Multiplier = @MarkupAsPercent / 100 + 1
7> SELECT Id, Name, salary,
8> salary * @Multiplier AS "Marked Up salary", "New salary" =
9> CASE WHEN FLOOR(salary * @Multiplier + .24) > FLOOR(salary * @Multiplier)
10> THEN FLOOR(salary * @Multiplier) + .95
11> WHEN FLOOR(salary * @Multiplier + .5) >
12> FLOOR(salary * @Multiplier)
13> THEN FLOOR(salary * @Multiplier) + .75
14> ELSE FLOOR(salary * @Multiplier) + .49
15> END
16> FROM Employee
17> GO
1>
2> EXEC spMarkupTest @MarkupAsPercent = 1.2
3> GO
Id Name salary Marked Up salary New salary
----------- ---------- ----------- --------------------- ----------------------
1 Jason 40420 40905.0400 40905.4900
2 Robert 14420 14593.0400 14593.4900
3 Celia 24020 24308.2400 24308.4900
4 Linda 40620 41107.4400 41107.4900
5 David 80026 80986.3120 80986.4900
6 James 70060 70900.7200 70900.7500
7 Alison 90620 91707.4400 91707.4900
8 Chris 26020 26332.2400 26332.4900
9 Mary 60020 60740.2400 60740.4900
(9 rows affected)
1>
2> drop table employee
3> GO
1>