HAJ*_*JAJ 15 sql t-sql sql-server sql-server-2008
我想使用case语句从函数中获取一个值.我尝试了以下但它不起作用:
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
Run Code Online (Sandbox Code Playgroud)
Mar*_*ers 17
CASE表达式有两种类型:简单和搜索.您必须选择其中一种 - 您不能在一个表达式中使用两种类型的混合物.
试试这个:
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
Run Code Online (Sandbox Code Playgroud)
此外,您需要将结果分配给其他人已经指出的内容.
ber*_*d_k 16
使用RETURN @Period时,必须为@Period赋值.以下示例显示如何构造代码,以便不需要声明局部变量.
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
Run Code Online (Sandbox Code Playgroud)
声明第二个变量,然后设置该值,因为您没有重置@Period.
例如:
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;
Run Code Online (Sandbox Code Playgroud)
我已经这样离开了,因为我假设您要为每个CASE语句更改您的值.