我有一个与此类似的 XML 文件(删除了更多节点和详细信息):

<?xml version="1.0" encoding="utf-8"?>
<Message xmlns="http://www.theia.org.uk/ILR/2011-12/1" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<Header>
    <CollectionDetails>
        <Collection>ILR</Collection>
        <Year>1112</Year>
        <FilePreparationDate>2011-10-06</FilePreparationDate>
    </CollectionDetails>
    <Source>
        <ProtectiveMarking>PROTECT-PRIVATE</ProtectiveMarking>          
    </Source>
</Header>
<SourceFiles>
    <SourceFile>
        <SourceFileName>A10004705001112004401.ER</SourceFileName>
        <FilePreparationDate>2011-10-05</FilePreparationDate>
    </SourceFile>
</SourceFiles>
<LearningProvider>
    <UKPRN>10004705</UKPRN>
    <UPIN>107949</UPIN>
</LearningProvider>
<Learner>
    <ULN>4682272097</ULN>
    <GivenNames>Peter</GivenNames>
    <LearningDelivery>
        <LearnAimRef>60000776</LearnAimRef>         
    </LearningDelivery>     
    <LearningDelivery>
        <LearnAimRef>ZPROG001</LearnAimRef>         
    </LearningDelivery>
</Learner>
<Learner>
    <ULN>3072094321</ULN>       
    <GivenNames>Thomas</GivenNames>     
    <LearningDelivery>
        <LearnAimRef>10055320</LearnAimRef>         
    </LearningDelivery>
    <LearningDelivery>
        <LearnAimRef>10002856</LearnAimRef>         
    </LearningDelivery>
    <LearningDelivery>
        <LearnAimRef>1000287X</LearnAimRef>         
    </LearningDelivery>
</Learner>
</Message>

我需要对此进行过滤，以便只有具有 ZPROG001 的子 LearningDelivery LearnAimRef 的学习者记录才会显示，因此在这种情况下输出将是第一个学习者而不是第二个:

<?xml version="1.0" encoding="utf-8"?>
<Message xmlns="http://www.theia.org.uk/ILR/2011-12/1" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<Header>
    <CollectionDetails>
        <Collection>ILR</Collection>
        <Year>1112</Year>
        <FilePreparationDate>2011-10-06</FilePreparationDate>
    </CollectionDetails>
    <Source>
        <ProtectiveMarking>PROTECT-PRIVATE</ProtectiveMarking>          
    </Source>
</Header>
<SourceFiles>
    <SourceFile>
        <SourceFileName>A10004705001112004401.ER</SourceFileName>
        <FilePreparationDate>2011-10-05</FilePreparationDate>
    </SourceFile>
</SourceFiles>
<LearningProvider>
    <UKPRN>10004705</UKPRN>
    <UPIN>107949</UPIN>
</LearningProvider>
<Learner>
    <ULN>4682272097</ULN>
    <GivenNames>Peter</GivenNames>
    <LearningDelivery>
        <LearnAimRef>60000776</LearnAimRef>         
    </LearningDelivery>     
    <LearningDelivery>
        <LearnAimRef>ZPROG001</LearnAimRef>         
    </LearningDelivery>
</Learner>
</Message>

我已经研究了如何执行此操作，并且认为执行此操作的正确方法是使用 XSL 转换来处理 xml 并根据需要输出到新文件(在 c# 中执行此操作)。在尝试围绕 XSLT 语法思考几个小时后，我仍然卡住了，无法获得我想要的输出。非常感谢任何帮助。

最佳答案

要复制大部分 XML 源文档，仅修改某些部分，您需要从身份转换开始。这只会复制所有内容。然后添加一个模板来覆盖 <Learner> 的身份模板您不想复制的元素:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
     xmlns:theia="http://www.theia.org.uk/ILR/2011-12/1">
  <!-- identity template -->
  <xsl:template match="@* | node()">
    <xsl:copy>
      <xsl:apply-templates select="@*|node()"/>
    </xsl:copy>
  </xsl:template>
  <!-- override the above template for certain Learner elements; output nothing. -->
  <xsl:template match="theia:Learner[
     not(theia:LearningDelivery/theia:LearnAimRef = 'ZPROG001')]">
  </xsl:template>
</xsl:stylesheet>

(从@andyb 借用命名空间前缀)。

关于xml - 根据子节点过滤 XML，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/7681577/