Reputation: 3787
I want to get one value from a function using a case statement. I tried the following but it does not work:
CREATE FUNCTION [FATMS].[fnReturnByPeriod]
(
@Period INT
)
RETURNS int
AS
BEGIN
SELECT CASE @Period
when 1 then 1
when @Period >1 and @Period <=7 then 1
when @Period >7 and @Period <=30 then 1
when @Period >30 and @Period<=90 then 1
when @Period >90 and @Period <=180 then 1
when @Period >180 and @Period <=360 then 1
else 0
END
RETURN @Period
END
Upvotes: 18
Views: 100044
Reputation: 1
CREATE FUNCTION TICKET_RATE (@RATE AS INT)
RETURNS VARCHAR (50)
AS
BEGIN
DECLARE @VISITED_PLACE VARCHAR (50)
IF @RATE > 400
SET @VISITED_PLACE = 'VISIT NEAR BY PLACE'
ELSE
RETURN
CASE
WHEN @RATE BETWEEN 0 AND 100 THEN 'VISIT MUMBAI'
WHEN @RATE BETWEEN 101 AND 200 THEN 'VISIT PUNE'
WHEN @RATE BETWEEN 201 AND 300 THEN 'VISIT DELHI'
WHEN @RATE BETWEEN 301 AND 400 THEN 'VISIT BENGLORE'
END
RETURN @VISITED_PLACE
END
Upvotes: 0
Reputation: 11966
When you use RETURN @Period, you must assign a value to @Period. The following example shows how to structure your code so that there is no need to declare a local variable.
CREATE FUNCTION [FATMS].[fnReturnByPeriod]
(
@Period INT
)
RETURNS INT
AS
BEGIN
RETURN
CASE
WHEN @Period = 1 THEN 1
WHEN @Period > 1 AND @Period <=7 THEN 1
WHEN @Period > 7 AND @Period <=30 THEN 1
WHEN @Period > 30 AND @Period<=90 THEN 1
WHEN @Period > 90 AND @Period <=180 THEN 1
WHEN @Period > 180 AND @Period <=360 THEN 1
ELSE 0
END
END
Upvotes: 19
Reputation: 107766
Here's a SET based approach to write your function in SQL Server 2008
CREATE FUNCTION [FATMS].[fnReturnByPeriod]
(
@Period INT
)
RETURNS int
AS
BEGIN
return isnull((SELECT min(v)
from (values
(1,1),
(7,2),
(30,3),
(90,4),
(180,5),
(360,6)) t(n,v)
where n >= @Period and @period>0), 0)
END
The way you have written it, all the CASE branches return 1, so you might as well use
return case when @period between 1 and 360 then 1 else 0 end
Upvotes: 1
Reputation: 16718
In what way does it "not work"? Does it produce an error when you try to execute it? Does it return an unexpected result?
Off the top of my head, there are several problems:
Your case statement returns 1
in all cases.
The case statement is also combining two different types of CASE
semantics:
CASE @Period
WHEN <condition> THEN <result>
WHEN <other condition> THEN <result>
ELSE <result>
END
or
CASE
WHEN @Period = <value> THEN <result>
WHEN @Period = <other value> THEN <result>
ELSE <result>
END
The second form allows you to use unrelated conditions, whereas the first can only check for different values of @Period
.
Further, you're returning the value of @Period, not the value generated by the CASE statement.
Upvotes: 1
Reputation: 48557
Declare a second variable and then set that value because you aren't resetting @Period
.
For example:
DECLARE @Output AS INT
SELECT @Output = CASE @Period
WHEN 1 then 1
WHEN @Period > 1 AND @Period <= 7 THEN 1 -- Should be 2
WHEN @Period > 7 AND @Period <= 30 THEN 1 -- Should be 3
WHEN @Period > 30 AND @Period<= 90 THEN 1 -- Should be 4
WHEN @Period > 90 AND @Period <= 180 THEN 1 -- Should be 5
WHEN @Period > 180 AND @Period <= 360 THEN 1 -- Should be 6
ELSE 0 END;
RETURN @Output;
I have left it like this as I'm assuming you are going to change your values for each of these CASE
statements.
Upvotes: 8
Reputation: 3829
You are not assigning @Period anywhere. Try with when @Period >1 and @Period <=7 then SET @Period = 1
and subsequently for all the other lines.
SQL does not have implicit returns and so you always have to set the variables before you return them.
Upvotes: 1
Reputation: 838716
There are two types of CASE expression: simple and searched. You must choose one or the other - you can't use a mixture both types in one expression.
Try this:
SELECT CASE
WHEN @Period = 1 THEN 1
WHEN @Period > 1 AND @Period <= 7 THEN 2
WHEN @Period > 7 AND @Period <= 30 then 3
-- etc...
ELSE 0
END
Also, you need to assign the result to something as others have already pointed out.
Upvotes: 19