1

我有如下 varchar 值

72,73,74

之后，我尝试将其拆分为逗号，然后将其转换为int以上值。比我想将Id 与我的User Table匹配。

CREATE FUNCTION Fn_MyFunction(@MyUserIdValues VARCHAR(100))  
RETURNS VARCHAR(300) AS  
BEGIN 

DECLARE @Result VARCHAR(300) = ''

Select UserName From UserTable
Where MyUserIdValues=UserIdValues


RETURN @Result

@Result 必须在一列中如下所示

Joe,Michael,Ricky

任何帮助将不胜感激。

谢谢。

4

6 回答 6

1

GROUP_CONCAT和FIND_IN_SET可能对您很方便。

尝试这个：

SELECT GROUP_CONCAT(UserName)
FROM UserTable 
WHERE FIND_IN_SET(MyUserIdValues,'72,73,74');
于 2013-10-03T06:24:19.323 回答
1

这样做的经典方法......

/*
create table Users
(
    id int,
    name nvarchar(max)
)

insert into Users
values 
(72, 'Joe'),
(73, 'Michael'),
(74, 'Ricky'),
(75, 'Manny'),
(76, 'Bob')

*/

CREATE FUNCTION dbo.Fn_MyFunction(@IdValues VARCHAR(100))  
RETURNS NVARCHAR(max) AS  
BEGIN 
DECLARE @Result NVARCHAR(max);
DECLARE @delimiter as nchar = ',';

    WHILE LEN(@IdValues) <> 0   
    BEGIN
        Declare @CurrentId int;
        If CHARINDEX(@delimiter, @IdValues) = 0  
        begin 
            Set @CurrentId = cast(@IdValues as int);
            Set @IdValues = ''
        End   
        Else  
        begin   
            Set @CurrentId = cast(left(@IdValues,  charindex(@delimiter, @IdValues) -1)  as int)
            Set @IdValues = Substring(@IdValues,  charindex(@delimiter, @IdValues) +1, len(@IdValues))   
        End  
        select @Result = Isnull(@Result + ',', '') + Isnull((Select Name From Users Where Id=@CurrentId),'(unknown)')
   END 

RETURN @Result
END
GO

Select dbo.Fn_MyFunction('72,73,74')
--Joe,Michael,Ricky
Select dbo.Fn_MyFunction('72,0,74')
--Joe,(unknown),Ricky
Select dbo.Fn_MyFunction('72,73,72,74,74')
--Joe,Michael,Joe,Ricky,Ricky
于 2013-10-03T06:26:58.357 回答
0

Check this out, it will solve your problem and it simple and perfect method

  Select Username from UserTable where MyUserIdValues IN ' + '('+ @Id +')'
于 2013-10-03T06:28:09.720 回答
0

你好，你也可以试试这个。

CREATE FUNCTION [DBO].[FN_SPLIT] ( @STRSTRING VARCHAR(MAX))
RETURNS @NAME TABLE (NAME VARCHAR(MAX))
AS
BEGIN
      DECLARE @USERID TABLE(ID INT)

      DECLARE @USERS TABLE (ID INT , NAME VARCHAR(50))
      INSERT INTO @USERS VALUES (72,'A'),(73,'B'),(74,'C')

      ;WITH STR_CTE(_START, _STOP) AS
      (
        SELECT  1, CHARINDEX(',' , @STRSTRING )
        UNION ALL
        SELECT  CAST(_STOP + 1 AS INT), CHARINDEX(',' ,@STRSTRING  , CAST((_STOP + 1) AS INT))
        FROM
            STR_CTE
        WHERE _STOP > 0
      )

      INSERT INTO @USERID (ID)     
      SELECT
            SUBSTRING(@STRSTRING , _START, CASE WHEN _STOP > 0 THEN _STOP -_START ELSE 4000 END) AS ID
      FROM STR_CTE

      DECLARE @STRNAME VARCHAR(MAX)
      SELECT
            @STRNAME = COALESCE(@STRNAME+',','') + U.NAME
      FROM
            @USERID UD
      INNER JOIN
            @USERS U ON UD.ID = U.ID

      INSERT INTO @NAME
      SELECT @STRNAME

      RETURN;
END
于 2013-10-03T15:27:18.003 回答
0

我找到了拆分字符串并将其插入表中的解决方案

GO
/****** Object:  UserDefinedFunction [dbo].[Split]    Script Date: 10/03/2013 11:45:16 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO

ALTER FUNCTION [dbo].[Split] (
      @InputString                  VARCHAR(8000),
      @Delimiter                    VARCHAR(50)
)

RETURNS @Items TABLE (
      Item                          int
)

AS
BEGIN
      IF @Delimiter = ' '
      BEGIN
            SET @Delimiter = ','
            SET @InputString = REPLACE(@InputString, ' ', @Delimiter)
      END

      IF (@Delimiter IS NULL OR @Delimiter = '')
            SET @Delimiter = ','



      DECLARE @Item                 VARCHAR(8000)
      DECLARE @ItemList       VARCHAR(8000)
      DECLARE @DelimIndex     INT

      SET @ItemList = @InputString
      SET @DelimIndex = CHARINDEX(@Delimiter, @ItemList, 0)
      WHILE (@DelimIndex != 0)
      BEGIN
            SET @Item = SUBSTRING(@ItemList, 0, @DelimIndex)
            INSERT INTO @Items VALUES (@Item)

            -- Set @ItemList = @ItemList minus one less item
            SET @ItemList = SUBSTRING(@ItemList, @DelimIndex+1, LEN(@ItemList)-@DelimIndex)
            SET @DelimIndex = CHARINDEX(@Delimiter, @ItemList, 0)
      END -- End WHILE

      IF @Item IS NOT NULL -- At least one delimiter was encountered in @InputString
      BEGIN
            SET @Item = @ItemList
            INSERT INTO @Items VALUES (@Item)
      END

      -- No delimiters were encountered in @InputString, so just return @InputString
      ELSE INSERT INTO @Items VALUES (cast(@InputString as int))

      RETURN

END -- End Function



Select UserName From UserTable where MyUserIdValues in(Select Item from Split('72,73,74',','))

//使用上次查询调用函数Split

于 2013-10-03T06:19:40.743 回答
0

我的尝试（我认为它更整洁）：

create function dbo.fn_UserNames(@ids varchar(max))
returns varchar(max) as
begin
    set @ids = @ids + ','

    declare @id table (Id int)
    while(@ids != '') begin
        insert into @id 
        select i
          from (select substring(@ids, 1, charindex(',', @ids, 0) - 1) i) a
         where i != ''

        if @ids like '%,%'
            set @ids = substring(@ids, charindex(',', @ids, 0) + 1, 
                       len(@ids) - charindex(',', @ids, 0))
        else
            set @ids = ''
    end

    declare @ret varchar(max)
    select @ret = isnull(@ret, '') + a.UserName + ','
      from adhoc.UserTable   a
      join @id               b  on a.UserId = b.Id

    return @ret
end
于 2013-10-03T06:42:18.657 回答