我有XML数据列,其中包含作为应用程序过程的一部分的问题和答案。我试图通过T-SQL / Dynamic SQL实现的是在有 目标 标记的任何地方派生唯一的路径集。因此,对于下面的xml示例,我希望得到类似
日志/客户端/客户端/部分/问题/ groupone /问题/目标日志/客户端/客户端/部分/问题/ grouptwo /问题/目标
然后的想法是使用它并遍历XML以得出所需标签的值。IE
[DATA] .value(’((/ log / clients / client / section / questions / groupone / question / target’,’NVARCHAR(MAX)’)
问题是每个应用程序都有不同的问题集和xml结构,即某些应用程序可能有更多问题,某些应用程序可能有不同的分组。但是,我想要的只是如果有一个标签,那么它的路径是什么。
我怎样才能最好地做到这一点?
<log> <clients> <client> <section name ="Apps鈥�> <questions> <groupone> <question> <target>Age</target> </question> <question> <target> Height</target> </question> <question> <target> Weight</target> </question> </groupone> <grouptwo name = "exercise"> <wording>what is your name</wording> <question> <id>1</id> <target>def<target> </question> </grouptwo> </questions> </section> </client> </clients> </log>
在这里,过时的方法FROM OPENXML可能是一个选择。检查此答案。
FROM OPENXML
在此链接中,您会发现一个不时发布的功能John Cappelletti,该功能将切碎所有XML(该功能代码下方的功劳)。
但我不确定,您真正想要实现的目标…为什么需要这条路?如果您对所有目标节点的值感兴趣,则可以执行以下操作(使用进行深度搜索//不需要确切的XPath)
//
XPath
SELECT t.value(N'(text())[1]','nvarchar(max)') FROM @xml.nodes('//target') AS A(t);
如果您真的需要 所有的一切 ,可以检查以下内容:
CREATE FUNCTION [dbo].[udf-XML-Hier](@XML xml) Returns Table As Return with cte0 as ( Select Lvl = 1 ,ID = Cast(1 as int) ,Pt = Cast(NULL as int) ,Element = x.value('local-name(.)','varchar(150)') ,Attribute = cast('' as varchar(150)) ,Value = x.value('text()[1]','varchar(max)') ,XPath = cast(concat(x.value('local-name(.)','varchar(max)'),'[' ,cast(Row_Number() Over(Order By (Select 1)) as int),']') as varchar(max)) ,Seq = cast(1000000+Row_Number() over(Order By (Select 1)) as varchar(max)) ,AttData = x.query('.') ,XMLData = x.query('*') From @XML.nodes('/*') a(x) Union All Select Lvl = p.Lvl + 1 ,ID = Cast( (Lvl + 1) * 1024 + (Row_Number() Over(Order By (Select 1)) * 2) as int ) * 10 ,Pt = p.ID ,Element = c.value('local-name(.)','varchar(150)') ,Attribute = cast('' as varchar(150)) ,Value = cast( c.value('text()[1]','varchar(max)') as varchar(max) ) ,XPath = cast(concat(p.XPath,'/',c.value('local-name(.)','varchar(max)'),'[',cast(Row_Number() Over(PARTITION BY c.value('local-name(.)','varchar(max)') Order By (Select 1)) as int),']') as varchar(max) ) ,Seq = cast(concat(p.Seq,' ',10000000+Cast( (Lvl + 1) * 1024 + (Row_Number() Over(Order By (Select 1)) * 2) as int ) * 10) as varchar(max)) ,AttData = c.query('.') ,XMLData = c.query('*') From cte0 p Cross Apply p.XMLData.nodes('*') b(c) ) , cte1 as ( Select R1 = Row_Number() over (Order By Seq),A.* From ( Select Lvl,ID,Pt,Element,Attribute,Value,XPath,Seq From cte0 Union All Select Lvl = p.Lvl+1 ,ID = p.ID + Row_Number() over (Order By (Select NULL)) ,Pt = p.ID ,Element = p.Element ,Attribute = x.value('local-name(.)','varchar(150)') ,Value = x.value('.','varchar(max)') ,XPath = p.XPath + '/@' + x.value('local-name(.)','varchar(max)') ,Seq = cast(concat(p.Seq,' ',10000000+p.ID + Row_Number() over (Order By (Select NULL)) ) as varchar(max)) From cte0 p Cross Apply AttData.nodes('/*/@*') a(x) ) A ) Select A.R1 ,R2 = IsNull((Select max(R1) From cte1 Where Seq Like A.Seq+'%'),A.R1) ,A.Lvl ,A.ID ,A.Pt ,A.Element ,A.Attribute ,A.XPath ,Title = Replicate('|---',Lvl-1)+Element+IIF(Attribute='','','@'+Attribute) ,A.Value From cte1 A /* Source: http://beyondrelational.com/modules/2/blogs/28/posts/10495/xquery-lab-58-select-from-xml.aspx Taken from John Cappelletti: https://stackoverflow.com/a/42729851/5089204 Declare @XML xml='<person><firstname preferred="Annie" nickname="BeBe">Annabelle</firstname><lastname>Smith</lastname></person>' Select * from [dbo].[udf-XML-Hier](@XML) Order by R1 */ GO DECLARE @xml XML= '<log> <clients> <client> <section name ="Apps"> <questions> <groupone> <question> <target>Age</target> </question> <question> <target> Height</target> </question> <question> <target> Weight</target> </question> </groupone> <grouptwo name = "exercise"> <wording>what is your name</wording> <question> <id>1</id> <target>def</target> </question> </grouptwo> </questions> </section> </client> </clients> </log>'; SELECT * FROM dbo.[udf-XML-Hier](@xml); GO