SQL Server/T-SQL/Math Functions/FLOOR — различия между версиями
Admin (обсуждение | вклад) м (1 версия) |
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(нет различий)
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Текущая версия на 10:20, 26 мая 2010
FLOOR(23.7)
1> SELECT FLOOR(23.7)
2> GO
-----
23
(1 rows affected)
1>
Floor a value
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> DECLARE @Markup money
3> DECLARE @Multiplier money
4>
5> SELECT @Markup = .10 -- Change the markup here
6> SELECT @Multiplier = @Markup + 1 -- We want the end price, not the amount
7> -- of the increase, so add 1
8> SELECT ID, Name, salary,
9> Salary * @Multiplier AS "Marked Up salary", "New salary" =
10> CASE WHEN FLOOR(salary * @Multiplier + .24) > FLOOR(salary * @Multiplier)
11> THEN FLOOR(salary * @Multiplier) + .95
12> WHEN FLOOR(salary * @Multiplier + .5) > FLOOR(salary * @Multiplier)
13> THEN FLOOR(salary * @Multiplier) + .75
14> ELSE FLOOR(salary * @Multiplier) + .49
15> END
16> FROM Employee
17> GO
ID Name salary Marked Up salary New salary
----------- ---------- ----------- --------------------- ----------------------
1 Jason 40420 44462.0000 44462.4900
2 Robert 14420 15862.0000 15862.4900
3 Celia 24020 26422.0000 26422.4900
4 Linda 40620 44682.0000 44682.4900
5 David 80026 88028.6000 88028.7500
6 James 70060 77066.0000 77066.4900
7 Alison 90620 99682.0000 99682.4900
8 Chris 26020 28622.0000 28622.4900
9 Mary 60020 66022.0000 66022.4900
(9 rows affected)
1>
2> drop table employee
3> GO
1>
FLOOR: return the largest integer less than the given value
1> -- FLOOR: return the largest integer less than the given value.
2>
3> SELECT FLOOR(-145.677)
4> GO
--------
-146
(1 rows affected)