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Question

我已经更新了代码的建议，但现在我得到一个新的错误标量函数：

--The function 
Create Function udf_GetCustomerAddress 
    (@CustomerID varchar) 
Returns varchar 
As 
Begin 
    declare @CustomerAddress varchar 

    select 
     @CustomerAddress = c.CompanyName, c.Address, c.City, 
     c.StateOrRegion, c.PostalCode, c.CustomerID 
    from 
     Customers c 
    where 
     c.CustomerID = @CustomerID 

    Return @CustomerAddress 
End 

再次感谢在：

Msg 141, Level 15, State 1, Procedure udf_GetCustomerAddress, Line 31
A SELECT statement that assigns a value to a variable must not be combined with data-retrieval operations.

我的代码粘贴下面提前！

Answer 1

您需要一个长度为所有varchar() s - 总是在SQL Server中使用长度。那么你可能想要连接结果。如果我假设所有类型都是字符串。 。 。

Create Function udf_GetCustomerAddress (
    @CustomerID varchar(255) 
) 
Returns varchar(max) 
As 
Begin 
    declare @CustomerAddress varchar(max); 

    select @CustomerAddress = (c.CompanyName + ' ' + 
           c.Address + ' ' + 
           c.City + ' ' + 
           c.StateOrRegion + ' ' + 
           c.PostalCode + ' ' + 
           c.CustomerID 
          ) 
    from Customers c 
    where c.CustomerID = @CustomerID; 

    Return @CustomerAddress; 
End; 

这假定你实际上想要所有的字段都带有一些分隔符。

但是，我不认为这是获得所需功能的最佳方式。最好在表格中添加一个计算列：

alter table Customers c 
    add CustomerAddress as (<expression here>); 

Answer 2

你应该RETURNS VARCHAR

Go 
--Address function 
Create Function udf_GetCustomerAddress 
(@CustomerID varchar) 
RETURNS varchar(max) 
As 
Begin 
    declare @CustomerAddress varchar(max); 

    select @CustomerAddress = concat(c.CompanyName, c.Address, c.City, 
    c.StateOrRegion, c.PostalCode, c.CustomerID) 
    from Customers c 
    where c.CustomerID = @CustomerID; 

    Return @CustomerAddress; 
End 

看一看CREATE FUNCTION语法改变Return varchar。