XSLT and xpath v1.0 find duplicates and aggregate

问题:

I wonder if there´s a way to search an xml for duplicates, then when found to aggregate all the duplicates to one node. F.eks

<car name="one">
 <person>john</person>
 <person>Jay</person>
</car>
<car name="two">
 <person>kim</person>
 <person>chris</person>
</car>
<car name="one">
 <person>jane</person>
 <person>liz</person>
</car>

Should be:

<car name="one">
 <person>john</person>
 <person>Jay</person>
 <person>jane</person>
 <person>liz</person>
</car>
<car name="two">
 <person>kim</person>
 <person>chris</person>
</car>

All help is much appreciated!

Br Kim

回答1:

In XSLT 1.0 you have to use something called the Muenchian Method which is a process of grouping elements using the key and generate-id functions.

The process goes something like this:

You first define a key that represents the data that you want to group with.

<xsl:key name="car-by-name" match="car" use="@name"/>

Then you you use that key in a template match by generating an id based on that key.

<xsl:apply-templates select="car[generate-id() = generate-id(key('car-by-name', @name)[1])]" mode="group"/>

Now that you have your nodes grouped, all you do is use that key again to grab all of the nodes within that key.

<xsl:apply-templates select="key('car-by-name', @name)"/>

So now to show it all together, when you take this XML (added document to make it well-formed).

<document>
  <car name="one">
    <person>john</person>
    <person>Jay</person>
  </car>
  <car name="two">
    <person>kim</person>
    <person>chris</person>
  </car>
  <car name="one">
    <person>jane</person>
    <person>liz</person>
  </car>
</document>

And apply this XSLT

<?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:key name="car-by-name" match="car" use="@name"/>

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

  <xsl:template match="document">
    <xsl:copy>
      <xsl:apply-templates select="car[generate-id() = generate-id(key('car-by-name', @name)[1])]" mode="group"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="car" mode="group">
    <car name="{@name}">
      <xsl:apply-templates select="key('car-by-name', @name)"/>
    </car>
  </xsl:template>

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

</xsl:stylesheet>

It produces this result.

<?xml version="1.0" encoding="UTF-8"?>
<document>
  <car name="one">
    <person>john</person>
    <person>Jay</person>

    <person>jane</person>
    <person>liz</person>
  </car>
  <car name="two">
    <person>kim</person>
    <person>chris</person>
  </car>
</document>

回答2:

If you are using XSLT 2.0 you can make use of xsl:for-each-group as follows:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>

    <xsl:template match="/data">
        <data>
            <xsl:for-each-group select="car" group-by="@name">
                <car name="{current-grouping-key()}">
                    <xsl:for-each select="current-group()/person">
                        <person><xsl:value-of select="." /></person>
                    </xsl:for-each>
                </car>
            </xsl:for-each-group>
        </data>
    </xsl:template>
</xsl:stylesheet>

When applied on the following input:

<?xml version="1.0" encoding="UTF-8"?>
<data>
    <car name="one">
        <person>john</person>
        <person>Jay</person>
    </car>
    <car name="two">
        <person>kim</person>
        <person>chris</person>
    </car>
    <car name="one">
        <person>jane</person>
        <person>liz</person>
    </car>
</data>

It produces the correct ouput:

<?xml version="1.0" encoding="UTF-8"?>
<data>
    <car name="one">
        <person>john</person>
        <person>Jay</person>
        <person>jane</person>
        <person>liz</person>
    </car>
    <car name="two">
        <person>kim</person>
        <person>chris</person>
    </car>
</data>