0

大家好我有以下存储过程

SELECT DISTINCT QuestionId, AnswerId, COUNT(AnswerId) AS Cntr,
    (SELECT     COUNT(AnswerId) AS ttl
     FROM          QUserAnswers
     WHERE      (QuestionId = QUAM.QuestionId)) AS TtlCnt
FROM         QUserAnswers AS QUAM
WHERE     (QuestionId IN (@QuestionIdIn))
GROUP BY QuestionId, AnswerId
ORDER BY QuestionId

@QuestionIdI我以格式传入' 但是,它在将 varchar 值转换为数据类型1,2,3,4,5'时抛出错误 Conversion failed 。'1,2,3,4,5,6'int

谁能给我一些指导来解决它

4

4 回答 4

1

正如 Tim Schelter 在他提供的链接中所建议的那样:-

首先,您需要创建一个解析输入的函数

 CREATE FUNCTION inputParser (@list nvarchar(MAX))
 RETURNS @tbl TABLE (number int NOT NULL) AS
 BEGIN
 DECLARE @pos        int,
       @nextpos    int,
       @valuelen   int

 SELECT @pos = 0, @nextpos = 1

 WHILE @nextpos > 0
 BEGIN
  SELECT @nextpos = charindex(',', @list, @pos + 1)
  SELECT @valuelen = CASE WHEN @nextpos > 0
                          THEN @nextpos
                          ELSE len(@list) + 1
                     END - @pos - 1
  INSERT @tbl (number)
     VALUES (convert(int, substring(@list, @pos + 1, @valuelen)))
  SELECT @pos = @nextpos
 END
 RETURN
END

然后在您的 SP 中使用该功能

 CREATE PROCEDURE usp_getQuestion
 @QuestionIdIn varchar(50) 
 AS
 Begin
 Select Distinct QuestionId, AnswerId, COUNT(AnswerId) AS Cntr,
    (SELECT     COUNT(AnswerId) AS ttl
      FROM   QUserAnswers
      WHERE      QuestionId = QUAM.QuestionId) as TtlCnt
 from QUserAnswers AS QUAM
 inner join inputParser (@QuestionIdIn) i ON QuaM.QuestionId = i.number
 GROUP BY QuestionId, AnswerId
 ORDER BY QuestionId
 End

EXEC usp_getQuestion '1, 2, 3, 4'
于 2012-07-26T12:22:56.583 回答
0

尝试在此处应用此逻辑。

Declare @str varchar(100)='''1'',''2'',''3'',''4'',''5'''
select @str
DECLARE @s int = 1
select * from emp where CAST(@s as varchar) in ('1','2','3','4','5')
于 2012-07-26T11:44:23.570 回答
0

代替

WHERE     (QuestionId IN (@QuestionIdIn)) 

利用

WHERE charindex(','+cast(@QuestionId as varchar(100))+',',','+@QuestionID+',')>0

如果您担心性能,请使用拆分功能并与主表连接

于 2012-07-26T11:26:31.523 回答
0

尝试改变条件:

WHERE     (QuestionId IN (@QuestionIdIn))

where (','+@QuestionIdIn+',' like '%,'+convert(varchar(50),QuestionId)+',%')

例如 @QuestionIdIn='1,2,3,4,5,6' 和 QuestionId=3 这意味着:

where (',1,2,3,4,5,6,' like '%,3,%')

如果 3 在此列表中,则为真。

于 2012-07-26T12:56:04.987 回答