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Performing a "Group By" query in XPath XSL

Given the following XML:

<results name="queryResults">
  <int name="intfield1:[* TO 10]">11</int> 
  <int name="intfield2:[10 TO 20]">9</int> 
  <int name="intfield1:[10 TO 20]">12</int> 
</results>

I would like to produce this XML:

<results>
    <field name="numberfield1">
        <value name="[* TO 10]">11</value>
        <value name="[10 TO 10]">12</value>
    </field>
    <field name="numberfield2">
        <value name="[10 TO 20]">9</value>
    </field>
</results>

I can't think how to do this in XSL mainly because i'm wanting to Group by the numbericfield.. All i can come up with is this:

<xsl:if test="count(results/int) &gt; 0">
    <results>
    <xsl:for-each select="results/int">
        <field>
            <xsl:attribute name="name">
                <xsl:value-of select="substring-before(@name, ':')"/></xsl:attribute>
            <value>
                <xsl:attribute name="name">
                    <xsl:value-of select="substring-after(@name, ':') "/>
                </xsl:attribute>
                <xsl:value-of select="."/>
            </value>
        </field>
    </xsl:for-each>
    </results>
</xsl:if>

However this doesn't produce the nice grouped list instead i get this:

<results>
    <field name="numberfield1">
        <value name="[* TO 10]">11</value>
    </field>
    <field name="numberfield2">
        <value name="[10 TO 20]">9</value>
    </field>
    <field name="numberfield1">
        <value name="[10 TO 10]">12</value>
    </field>
</results>

If someone can stear me in the right direction.. That would be great?

Thanks

like image 666
CraftyFella Avatar asked Dec 18 '09 16:12

CraftyFella


1 Answers

To do this in XSLT 1.0, you will have to use a technique called "muenchian grouping". First create a key of the nodes on which you wish to group

<xsl:key name="intfield" match="int" use="substring-before(@name, ':')" />

Next, you iterate it through all the nodes, but only selecting the ones that happen to be first in the relevant group

<xsl:for-each select="int[generate-id() = generate-id(key('intfield', substring-before(@name, ':'))[1])]">

Next, you can iterate use the key to iterate over all nodes in the group

<xsl:variable name="intfieldname" select="substring-before(@name, ':')"/>
<xsl:for-each select="key('intfield', $intfieldname)">

Putting this all together gives

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
   <xsl:output method="xml"/>
   <xsl:key name="intfield" match="int" use="substring-before(@name, ':')"/>
   <xsl:template match="/results">
      <results>
         <xsl:for-each select="int[generate-id() = generate-id(key('intfield', substring-before(@name, ':'))[1])]">
            <xsl:variable name="intfieldname" select="substring-before(@name, ':')"/>
            <field>
               <xsl:attribute name="name">
                  <xsl:value-of select="$intfieldname"/>
               </xsl:attribute>
               <xsl:for-each select="key('intfield', $intfieldname)">
                  <value>
                     <xsl:attribute name="name">
                        <xsl:value-of select="substring-after(@name, ':')"/>
                     </xsl:attribute>
                     <xsl:value-of select="."/>
                  </value>
               </xsl:for-each>
            </field>
         </xsl:for-each>
      </results>
   </xsl:template>
</xsl:stylesheet>

In your example, 'intfield' becomes 'numberfield' though. I have kept the name as 'intfield' in the above example.

  • fixed typo.
like image 103
Tim C Avatar answered Sep 27 '22 02:09

Tim C