是否有一种低成本的方式来查询具有嵌套集模型的表,以便从特定级别获取节点集合?
即如何从我的树级别 n 中获取 IEnumerable?即如何从我的树级别 n-1 获得 IEnumerable?
谢谢!
是否有一种低成本的方式来查询具有嵌套集模型的表,以便从特定级别获取节点集合?
即如何从我的树级别 n 中获取 IEnumerable?即如何从我的树级别 n-1 获得 IEnumerable?
谢谢!
也许比你需要的多一点,但可以减少。
以下 TVF 将解析几乎任何 XML 结构并返回规范化的层次结构。
作为 TVF,您可以应用任何所需WHERE
的,例如WHERE Lvl=3
,甚至在CROSS APPLY
例子
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
退货
有兴趣的TVF
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(10000001 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
*/