Use XSLT/XPATH to select elements having a child e

问题:

I have a large (200k lines) XML file as a report from a tool (reporting on a VSS database). It consists of a large number of <file> elements like this:

<file>
 <name>file.bat</name>
 <version>111</version>
 <checkedout>No</checkedout>
 <binary>Text</binary>
 <vss_path>$/Code/file.bat</vss_path>
 <original_path>C:\code\file.bat</original_path>
 <action>Labeled &apos;1.2.3.4&apos;</action>
 <date>27/09/2013 09:08:00</date>
 <comment></comment>
 <label>1.2.3.4</label>
 <label_comment></label_comment>
 <user>John</user>
 <shared_links>
  <shared_link>$/Beta_1</shared_link>
  <shared_link>$/Branches/New_Feature</shared_link>
 </shared_links>
</file>

I want to find only the <file> elements which have at least one <shared_link> starting with/prefixed by "$/Beta".

In an ideal world, all I want for each matching element are the <name>, <vss_path> and (matching) <shared_link> parts, but that's not exactly important.

I'm not well-versed in XSLT/XPATH but believe those can do something like this?

回答1:

This XSLT stylesheet:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>

  <xsl:template match="node()|@*">
    <xsl:copy>
      <xsl:apply-templates select="node()|@*"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="root">
    <xsl:copy>
      <xsl:apply-templates select="file[shared_links[shared_link[starts-with(., '$/Beta')]]]"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="file">
    <xsl:copy>
      <xsl:apply-templates select="name | vss_path | shared_links"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="shared_links">
    <xsl:copy>
      <xsl:apply-templates select="shared_link[starts-with(., '$/Beta')]"/>
    </xsl:copy>
  </xsl:template>

</xsl:stylesheet>

when applied to this input XML (as yours but with an extra, non-matching file added):

<root>
  <file>
    <name>file.bat</name>
    <version>111</version>
    <checkedout>No</checkedout>
    <binary>Text</binary>
    <vss_path>$/Code/file.bat</vss_path>
    <original_path>C:\code\file.bat</original_path>
    <action>Labeled &apos;1.2.3.4&apos;</action>
    <date>27/09/2013 09:08:00</date>
    <comment></comment>
    <label>1.2.3.4</label>
    <label_comment></label_comment>
    <user>John</user>
    <shared_links>
      <shared_link>$/Alpha_1</shared_link>
      <shared_link>$/Branches/New_Feature</shared_link>
    </shared_links>
  </file>
  <file>
    <name>file.bat</name>
    <version>111</version>
    <checkedout>No</checkedout>
    <binary>Text</binary>
    <vss_path>$/Code/file.bat</vss_path>
    <original_path>C:\code\file.bat</original_path>
    <action>Labeled &apos;1.2.3.4&apos;</action>
    <date>27/09/2013 09:08:00</date>
    <comment></comment>
    <label>1.2.3.4</label>
    <label_comment></label_comment>
    <user>John</user>
    <shared_links>
      <shared_link>$/Beta_1</shared_link>
      <shared_link>$/Branches/New_Feature</shared_link>
    </shared_links>
  </file>
</root>

produces the following output XML:

<root>
  <file>
    <name>file.bat</name>
    <vss_path>$/Code/file.bat</vss_path>
    <shared_links>
      <shared_link>$/Beta_1</shared_link>
    </shared_links>
  </file>
</root>

回答2:

Use

<xsl:template match="/">
  <xsl:apply-templates select="//file[shared_links/shared_link[starts-with(., '$/Beta')]]"/>
</xsl:template>

<xsl:template match="file">
  <xsl:copy-of select="name | vss_path | shared_links/shared_link"/>
</xsl:template>

to output those elements. That way however the result is an XML fragment with multiple top level elements, if you want an XML document then change the first template to

<xsl:template match="/">
  <root>
  <xsl:apply-templates select="//file[shared_links/shared_link[starts-with(., '$/Beta')]]"/>
  </root>
</xsl:template>