如果在XSLT中，最好的方法是使用每个内部

Question

我将以下XML数据作为我的XSLT的输入：

<?xml version="1.0" encoding="UTF-8"?>
<Application>
    <Data>
        <Data1>
            <name>Michale</name>
            <age>65</age>
            <Info>
                <Alias name="M">
                    <Contactmail>[email protected]</Contactmail>
                    <ContactPh>8988900009</ContactPh>
                </Alias>

                <Alias name="Q">
                    <Contactmail>[email protected]</Contactmail>
                    <ContactPh>8988900009</ContactPh>
                </Alias>

            </Info>
        </Data1>
        <Data1>
            <name>Albert</name>
            <age>69</age>
            <Info>
                <Alias name="A">
                    <Contactmail>[email protected]</Contactmail>
                    <ContactPh>89889908709</ContactPh>
                </Alias>

                <Alias name="P">
                    <Contactmail>[email protected]</Contactmail>
                    <ContactPh>8988988779</ContactPh>
                </Alias>
            </Info>
        </Data1>
    </Data>
</Application>

我想通过其别名与“M”匹配的Data1块，即：

      <Application>
         <Data>
           <Data1>
              <name>Michale</name>
              <age>65</age>
              <Info>
                <Alias name=M>
                   <Contactmail>[email protected]</Contactmail>
                   <ContactPh>8988900009</ContactPh>
                </Alias> 
                <Alias name=Q>
                   <Contactmail>[email protected]</Contactmail>
                   <ContactPh>8988900009</ContactPh>
                </Alias>  
              </Info>
           </Data1>
       </Data>
     </Application>

我被困在如何在测试条件下访问循环（即Alias）？有没有更好的方法来做这个xslt？

<xsl:for-each select="./*[local-name() = 'Application']/*[local-name() = 'Data']">
   <xsl:if test="">
   ....
   </xsl:if>
</xsl:for-each>

Answer 1

以下模板将完成这项工作。解释在代码中。

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml" indent="yes"/>
  <xsl:strip-space elements="*" />                         <!-- Removes unnecessary space between elements -->

  <!-- identity template -->                               <!-- Copies all nodes not matched by other templates -->
  <xsl:template match="node()|@*">
    <xsl:copy>
      <xsl:apply-templates select="node()|@*" />
    </xsl:copy>
   </xsl:template>  

  <xsl:template match="Data1[Info/Alias/@name != 'M']" />  <!-- Ignores all Data1 elements which don't have an @name='M' attribute child -->
  <xsl:template match="Data1[Info/Alias/@name = 'M']">     <!-- Matches all Data1 elements which have the desired child attribute -->
    <xsl:copy>
      <xsl:apply-templates select="node()|@*" />
    </xsl:copy>
  </xsl:template>

</xsl:stylesheet>

它的输出是：

<?xml version="1.0"?>
<Application>
  <Data>
    <Data1>
      <name>Michale</name>
      <age>65</age>
      <Info>
        <Alias name="M">
          <Contactmail>[email protected]</Contactmail>
          <ContactPh>8988900009</ContactPh>
        </Alias>
        <Alias name="Q">
          <Contactmail>[email protected]</Contactmail>
          <ContactPh>8988900009</ContactPh>
        </Alias>
      </Info>
    </Data1>
  </Data>
</Application>

Answer 2

<xsl:strip-space elements="*"/>
    <xsl:template match="@*|node()">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="Application">
        <xsl:copy>
            <xsl:for-each select="Data/Data1">
                <xsl:if test="Info/Alias[@name='M']">
                <Data>
                    <Data1>

                        <xsl:apply-templates/>

                    </Data1>
                </Data>
                </xsl:if>
            </xsl:for-each>
        </xsl:copy>
    </xsl:template>
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