问题:如何在多个函数之间创建序列?
我有各种创建xml数据的函数,每个函数都可以创建多个"Party“节点集。所有函数都从同一个父节点开始。我希望输出如下所示,其中每一方,不管它来自哪个函数,都有连续的序列号。期望产出:
<PARTIES>
<PARTY SequenceNumber="1" label="PARTY_1">
...
<PARTY SequenceNumber="2" label="PARTY_2">
...
<PARTY SequenceNumber="3" label="PARTY_3">
...
</PARTIES>现在,我通过一个返回xml的函数输出我的xml,并且我想要排序的函数被分组在PARTIES节点下面:
SELECT [dbo].[GetFunction1Xml](@Id),
[dbo].[GetFunction2Xml](@Id),
[dbo].[GetFunction3Xml](@Id)
FOR XML PATH(''), ROOT('PARTIES'), TYPE每个函数从不同的地方收集信息,如下所示:
ALTER GetFunction1XML
...
RETURNS XML (
SELECT [label] = 'PARTY_' + CONVERT(NVARCHAR,ROW_NUMBER() OVER(ORDER BY (SELECT NULL)))
[Var1] = ....,
[Var2] = ....,
FROM [Table]
FOR XML PATH('PARTY'), TYPE)
END;我试图使用一个序列,但是在用户定义的函数中是不允许的。
CREATE SEQUENCE Party_Seq
AS INTEGER
START WITH 1
INCREMENT BY 1
MINVALUE 1
NO CYCLE; 我还在每个函数中尝试了以下内容,因为如果要在同一个函数中有两个当事方由UNION连接,它就可以工作。然而,由于各方都有不同的功能,所以每次都重新启动PARTY_1。
SELECT [@label] = 'PARTY_' + CONVERT(NVARCHAR,ROW_NUMBER() OVER(ORDER BY (SELECT NULL)))因此,例如,如果我用一个通用函数替换2个函数,它看起来会是这样,它正确地打印出信息;但是,我有太多的函数可以完成这个任务。
ALTER GetGenericFunctionXML
...
RETURNS XML (
SELECT [@seq] = ROW_NUMBER() OVER(ORDER BY (SELECT NULL))
[@label] = 'PARTY_' + CONVERT(NVARCHAR,ROW_NUMBER() OVER(ORDER BY (SELECT NULL)))
[Var1] = [food].[fruit],
[Var2] = [food].[meat]
FROM ( SELECT 'Apple' AS [fruit],
'Bacon' AS [meat]
FROM [Table1]
UNION ALL
SELECT 'Grape',
'Pork'
FROM [Table2]
) AS [food]
FOR XML PATH('PARTY'), TYPE)
END;输出:
<PARTIES>
<PARTY SequenceNumber="1" label="PARTY_1">
<Var1>Apple</Var1>
<Var2>Bacon</Var2>
<PARTY SequenceNumber="2" label="PARTY_2">
<Var1>Grape</Var1>
<Var2>Pork</Var2>
<PARTY SequenceNumber="3" label="PARTY_3">
</PARTIES>我也尝试将一个参数传递给函数,但是由于它们是函数,所以它们不能输出值(我相信只有存储过程才能这样做。如果我错了,请纠正我。)
发布于 2016-08-31 13:53:43
您可以用FLWOR来解决这个问题。
CREATE FUNCTION dbo.f1() RETURNS XML AS
BEGIN
RETURN
'<PARTY label="PARTY_f1a">
<Var1>f1a.1</Var1>
<Var2>f1a.2</Var2>
</PARTY>
<PARTY label="PARTY_f1b">
<Var1>f1b.1</Var1>
<Var2>f1b.2</Var2>
</PARTY>
<PARTY label="PARTY_f1c">
<Var1>f1c.1</Var1>
<Var2>f1c.2</Var2>
</PARTY>';
END
GO
CREATE FUNCTION dbo.f2() RETURNS XML AS
BEGIN
RETURN
'<PARTY label="PARTY_f2a">
<Var1>f2a.1</Var1>
<Var2>f2a.2</Var2>
</PARTY>
<PARTY label="PARTY_f2b">
<Var1>f2b.1</Var1>
<Var2>f2b.2</Var2>
</PARTY>';
END
GO--查询从这里开始
WITH AllPartyNodes AS
(
SELECT
(
SELECT dbo.f1()
,dbo.f2()
FOR XML PATH(''),TYPE
) AS AllTogether
)
,NumberedSequences AS
(
SELECT ROW_NUMBER() OVER(ORDER BY (SELECT NULL)) AS SequenceNr
,The.Party.query('.') AS TheNode
FROM AllPartyNodes
CROSS APPLY AllTogether.nodes('/PARTY') AS The(Party)
)
SELECT TheNode.query('let $p:=/PARTY[1]
let $lbl:=$p/@label
let $nr:=sql:column("SequenceNr")
return
<PARTY seq="{$nr}" label="{$lbl}" >
{$p/*}
</PARTY>'
) AS [node()]
FROM NumberedSequences
FOR XML PATH(''),ROOT('PARTIES')
GO
DROP FUNCTION dbo.f1;
DROP FUNCTION dbo.f2;更新另一种方法
您可以提取数据并重新构建它。
把这个放在我的"NumberedSequence“CTE下面
,TheData AS
(
SELECT *
,TheNode.value('(PARTY/@label)[1]','nvarchar(max)') AS Label
,TheNode.query('PARTY/*') AS InnerNodes
FROM NumberedSequences
)
SELECT SequenceNr AS [@seq]
,Label AS [@label]
,InnerNodes AS [node()]
FROM TheData
FOR XML PATH('PARTY'),ROOT('PARTIES')更新2
与主查询的函数相同。
CREATE FUNCTION dbo.f1() RETURNS XML AS
BEGIN
RETURN
'<PARTY label="PARTY_f1a">
<Var1>f1a.1</Var1>
<Var2>f1a.2</Var2>
</PARTY>
<PARTY label="PARTY_f1b">
<Var1>f1b.1</Var1>
<Var2>f1b.2</Var2>
</PARTY>
<PARTY label="PARTY_f1c">
<Var1>f1c.1</Var1>
<Var2>f1c.2</Var2>
</PARTY>';
END
GO
CREATE FUNCTION dbo.f2() RETURNS XML AS
BEGIN
RETURN
'<PARTY label="PARTY_f2a">
<Var1>f2a.1</Var1>
<Var2>f2a.2</Var2>
</PARTY>
<PARTY label="PARTY_f2b">
<Var1>f2b.1</Var1>
<Var2>f2b.2</Var2>
</PARTY>';
END
GO
--The main query as function
CREATE FUNCTION dbo.f3() RETURNS XML AS
BEGIN
DECLARE @Result XML;
WITH AllPartyNodes AS
(
SELECT
(
SELECT dbo.f1()
,dbo.f2()
FOR XML PATH(''),TYPE
) AS AllTogether
)
,NumberedSequences AS
(
SELECT ROW_NUMBER() OVER(ORDER BY (SELECT NULL)) AS SequenceNr
,The.Party.query('.') AS TheNode
FROM AllPartyNodes
CROSS APPLY AllTogether.nodes('/PARTY') AS The(Party)
)
SELECT @Result=
(
SELECT TheNode.query('let $p:=/PARTY[1]
let $lbl:=$p/@label
let $nr:=sql:column("SequenceNr")
return
<PARTY seq="{$nr}" label="{$lbl}" >
{$p/*}
</PARTY>'
) AS [node()]
FROM NumberedSequences
FOR XML PATH(''),ROOT('PARTIES'), TYPE
)
RETURN @Result;
END
GO
SELECT dbo.f3();
GO
DROP FUNCTION dbo.f1;
DROP FUNCTION dbo.f2;
DROP FUNCTION dbo.f3;https://stackoverflow.com/questions/39238877
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