UDF在SQL SELECT语句中找不到

时间:2018-01-30 06:37:01

标签: sql sql-server select user-defined-functions

我无法理解从以下链接引用的选择UDF语句:

How to split a comma-separated value to columns

CREATE FUNCTION dbo.Split (
  @InputString                  VARCHAR(500),
  @Delimiter                    VARCHAR(1)
)

RETURNS @Items TABLE (
  Item                          VARCHAR(500)
)

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

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

--INSERT INTO @Items VALUES (@Delimiter) -- Diagnostic
--INSERT INTO @Items VALUES (@InputString) -- Diagnostic

  DECLARE @Item           VARCHAR(500)
  DECLARE @ItemList       VARCHAR(500)
  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 (@InputString)

  RETURN

END -- End Function
GO

我检查了权限,并确保使用以下命令在正确的模式中创建了该函数:

SELECT * FROM INFORMATION_SCHEMA.ROUTINES

但是,当我运行以下内容时,我遇到了问题:

select dbo.Split(TargetFolderId, ',') from ReportConfig where ReportId = 9

错误:

  

找不到列“dbo”或用户定义的函数或   聚合“dbo.Split”,或名称不明确。

但我能做到

select * from dbo.Split('stringA,stringB', ',')

我们俩正在使用的SSMS版本之间是否存在差异?

我正在使用SQL2017 - SSMS v17.4。

2 个答案:

答案 0 :(得分:2)

显然,该函数被定义为表值,你不能像那样使用它

select dbo.Split(TargetFolderId, ',') from ReportConfig where ReportId = 9

但您可以执行CROSS APPLY来获得TargetFolderId分割值。

select sp.Item from ReportConfig rc
CROSS APPLY dbo.Split(rc.TargetFolderId) AS sp
where rc.ReportId = 9

答案 1 :(得分:0)

这是你的答案

RETURNS @Items TABLE (
  Item                          VARCHAR(500)
)

在你的代码中,你将返回一个表,所以你必须使用类似这样的函数

Select * FROM dbo.Split('1,2,3,4,5,6',',')

结果

Item
1
2
3
4
5
6