XML列--三级层次结构--具有交叉应用功能

Question

我之前的问题已经解决了 此处. 现在我又给它增加了一个复杂程度--数据是嵌套的父、子、孙。

你可以在这里看到和运行示例。https:/dbfiddle.uk?rdbms=sqlserver_2019&fiddle=df2766c95383d4c8c2d1f55539634341。

示例代码，其中Leg1可能是出去的行程，Leg2可能是回来的行程。每个腿可以有一个或多个航班。

DECLARE @xml XML='
<Reservation>
  <Name>Neal</Name>
    <Leg seq=''1''>
      <Flight>12</Flight>
    </Leg>
    <Leg seq=''2''>
      <Flight>34</Flight>
      <Flight>56</Flight>
    </Leg>
</Reservation>'
select @xml

DECLARE @xmlTable TABLE (
    xmlDoc Xml
);
Insert into @xmltable values (@xml)
--Select xmlDoc from @XmlTable 

Select xmlDoc.value('(//Name)[1]', 'varchar(30)') as Passenger,
       XmlData2.xmlDoc2.query('.') as XmlData2,
       XmlData2.xmlDoc2.value('./@seq', 'int') as LegSeq,
       XmlData3.xmlDoc3.query('.') as XmlData3,
       XmlData3.xmlDoc3.value('.', 'varchar(20)') as Flight
FROM @xmlTable as t
     CROSS APPLY 
        t.xmlDoc.nodes('//Leg') AS XmlData2(xmlDoc2)
     CROSS APPLY 
        t.xmlDoc.nodes('//Flight') AS XmlData3(xmlDoc3)

问题是我仍然需要3行返回，但现在我得到了6行。

预期的结果应该是。

Neal LegSeq=1 Flight=12 
Neal LegSeq=2 Flight=34
Neal LegSeq=2 Flight=56

Microsoft SQL Server 2019 (RTM) - 15.0.2000.5 (X64) Sep 24 2019 13:48:23 Copyright (C) 2019 Microsoft Corporation Developer Edition (64-bit) on Windows Server 2019 Standard 10.0 (Build 17763: )

Answer 1

在第二个apply中，你想应用到来自于以下的节点 XmlData2.xmlDoc2. 按照你的写法，它将再次从根部寻找节点，这将适用于所有的 Flight XML中的元素。

DECLARE @xml XML='
<Reservation>
  <Name>Neal</Name>
    <Leg seq=''1''>
      <Flight>12</Flight>
    </Leg>
    <Leg seq=''2''>
      <Flight>34</Flight>
      <Flight>56</Flight>
    </Leg>
</Reservation>'
select @xml

DECLARE @xmlTable TABLE (
    xmlDoc Xml
);
Insert into @xmltable values (@xml)
--Select xmlDoc from @XmlTable 

Select xmlDoc.value('(//Name)[1]', 'varchar(30)') as Passenger,
       XmlData2.xmlDoc2.query('.') as XmlData2,
       XmlData2.xmlDoc2.value('./@seq', 'int') as LegSeq,
       XmlData3.xmlDoc3.query('.') as XmlData3,
       XmlData3.xmlDoc3.value('.', 'varchar(20)') as Flight
FROM @xmlTable as t
     CROSS APPLY 
        t.xmlDoc.nodes('//Leg') AS XmlData2(xmlDoc2)
     CROSS APPLY 
        XmlData2.xmlDoc2.nodes('Flight') AS XmlData3(xmlDoc3);