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我正在尝试编写一个函数来支持一些（旧版）SQL 代码，函数将用作查询的一部分：

SELECT
  q.*,
  fn_qncatxml(q.number, q.version) AS categories --XML aggregated category hierarchy
FROM
  [dbo].[qn] q

不幸的是，以下方法不起作用，因为 SQL Server 在函数中的 CTE 表达式存在一些已知问题（WITH 解决方案之前的分号）。不幸的是，我不确定如何将此解决方案应用于返回标量的函数，如下所示：

CREATE FUNCTION [dbo].[fn_qncatxml](@qnnumber INT, @qnversion INT)
RETURNS XML
WITH RETURNS NULL ON NULL INPUT
BEGIN
  RETURN
  (
    WITH [categories] AS
    (
      SELECT [qn].[number], [qn].[version], [cat].[catnumber], [qncat].[itemnumber], [cat].[parent], [cat].[description], 0 AS [distance]
      FROM [qn]
      LEFT JOIN [qncat] ON [qncat].[qnnumber] = [qn].[number] AND [qncat].[qnversion] = [qn].[version]
      LEFT JOIN [cat] ON [cat].[catnumber] = [qncat].[catnumber] AND [cat].[status] = '1'
      WHERE [qn].[number] = @qnnumber AND [qn].[version] = @qnversion

      UNION ALL

      SELECT [categories].[number], [categories].[version], [cat].[catnumber], [categories].[itemnumber], [cat].[parent], [cat].[description], [categories].[distance] + 1 AS [distance]
      FROM [categories]
      JOIN [cat] ON [cat].[catnumber] = [categories].[parent]
      WHERE
        [cat].[status] = '1'
    )
    SELECT DISTINCT * FROM [categories] FOR XML PATH('')
  )
END

当我将去掉 WITH 和 FOR XML PATH('') 之间的一部分，给出一些有效的参数并执行这个块时，它可以完美地工作。

有谁知道如何使这个 CTE 函数语法在我的场景中工作？

4

1 回答 1

3 
CREATE FUNCTION [dbo].[fn_qncatxml](@qnnumber INT, @qnversion INT)
RETURNS XML
WITH RETURNS NULL ON NULL INPUT
BEGIN
  DECLARE @Ret xml;

  WITH [categories] AS
  (
    SELECT [qn].[number], [qn].[version], [cat].[catnumber], [qncat].[itemnumber], [cat].[parent], [cat].[description], 0 AS [distance]
    FROM [qn]
    LEFT JOIN [qncat] ON [qncat].[qnnumber] = [qn].[number] AND [qncat].[qnversion] = [qn].[version]
    LEFT JOIN [cat] ON [cat].[catnumber] = [qncat].[catnumber] AND [cat].[status] = '1'
    WHERE [qn].[number] = @qnnumber AND [qn].[version] = @qnversion

    UNION ALL

    SELECT [categories].[number], [categories].[version], [cat].[catnumber], [categories].[itemnumber], [cat].[parent], [cat].[description], [categories].[distance] + 1 AS [distance]
    FROM [categories]
    JOIN [cat] ON [cat].[catnumber] = [categories].[parent]
    WHERE
      [cat].[status] = '1'
  )
  SELECT @Ret = (SELECT DISTINCT * FROM [categories] FOR XML PATH(''))

  RETURN @Ret
END
于 2011-07-14T13:48:50.250 回答