任何人都可以幫我找到代碼中的錯誤

Question

我寫了函數來獲取姓氏，憑據，後綴和中間名。我用名字取代了所有這些，以獲得名字。我將FirstName作爲null放在FirstName中。任何人都可以幫我解決它。

alter function FirstName(@Name varchar(100)) 
returns varchar(25) 
as 
begin 
declare @FirstName varchar(70) 
declare @LastName varchar(100) 
declare @Suffix varchar(100) 
declare @Cred1 varchar(100) 
declare @Cred2 varchar(100) 
declare @MiddleName varchar(100) 
set @LastName = dbo.LastName(@Name) 
set @Suffix = dbo.Suffix(@Name) 
set @Cred1 = dbo.Cred(@Name) 
set @Cred2 = dbo.Cred2(@Name) 
set @MiddleName = dbo.MiddleName(@Name) 
set @FirstName = @Name 
set @FirstName =case when len(@Cred1) >0 
      then stuff(@FirstName,charindex('%'+@Cred1+'%',@Name),len(@Cred1),' ') 
      else @FirstName 
      end 
set @FirstName =case when len(@Cred2) >0 
      then stuff(@Name,charindex('%'+@Cred2+'%',@Name),len(@Cred2),' ') 
      else @FirstName 
      end 
set @FirstName =case when len(@Suffix) >0 
      then stuff(@Name,charindex('%'+@Suffix+'%',@Name),len(@Suffix),' ') 
      else @FirstName 
      end 
set @FirstName =case when len(@LastName) >0 
      then stuff(@FirstName,charindex('%'+@LastName+'%',@Name),len(@LastName),' ') 
      else @FirstName 
      end 
set @FirstName =case when len(@MiddleName) >0 
      then stuff(@Name,charindex('%'+@MiddleName+'%',@Name),len(@MiddleName),' ') 
      else @FirstName 
      end 

return @FirstName 

end 

Answer 1

如果charindex('%'+@MiddleName+'%',@Name)返回一個負值或值比所述字段的長度它正試圖操作上，例如長@FirstName或@Name，則返回空字符串。

我會檢查你正在使用的值，看看是否是這種情況。

查看http://msdn.microsoft.com/en-us/library/ms188043.aspx查看該功能的詳細信息。