Re: [xsl] Grouping and Numbering

Subject: Re: [xsl] Grouping and Numbering
From: "Mukul Gandhi" <gandhi.mukul@xxxxxxxxx>
Date: Fri, 25 May 2007 22:24:30 +0530
I find this problem rather difficult to solve with pure XSLT 1.0. This
is due to the fact, that XSLT doesn't allow to maintain variable
state. If you are willing to write extension functions (for e.g. in
Java), it's easier to solve this problem.

Please consider the following example.

Java extension class:

public class Iterator
{
  private static int i = 0;

  public static int next()
  {
    i++;
    return i;
  }
}

XSLT stylesheet:

<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
                      xmlns:java="http://xml.apache.org/xalan/java";
                      exclude-result-prefixes="java"
                      version="1.0">

<xsl:output method="xml" indent="yes" />

<xsl:key name="by-group" match="result" use="details/group_id" />

 <xsl:template match="sample">
   <sample>
     <xsl:for-each select="result[generate-id() =
generate-id(key('by-group', details/group_id)[1])]">
       <xsl:sort select="details/group_id" data-type="number" />
       <xsl:variable name="x" select="position()" />
       <xsl:for-each select="key('by-group', details/group_id)">
         <output row="{java:Iterator.next()}"><xsl:value-of
select="details/group_id" /></output>
       </xsl:for-each>
       <xsl:if test="$x != last()">
         <output row="{java:Iterator.next()}" />
       </xsl:if>
     </xsl:for-each>
   </sample>
 </xsl:template>

</xsl:stylesheet>

Using Xalan-J (ver 2.7.0), when the above stylesheet is applied to the XML:

<sample>
  <result>
    <details>
      <group_id>250</group_id>
    </details>
  </result>
  <result>
    <details>
      <group_id>300</group_id>
    </details>
  </result>
  <result>
    <details>
      <group_id>250</group_id>
    </details>
  </result>
</sample>

The output produced is:

<?xml version="1.0" encoding="UTF-8"?>
<sample>
 <output row="1">250</output>
 <output row="2">250</output>
 <output row="3"/>
 <output row="4">300</output>
</sample>

PS: The stylesheet uses Muenchian method to do grouping. I am also
keen to know, is there any way, this problem can be solved with pure
XSLT 1.0.

On 5/25/07, Carl Radley <Carl.Radley@xxxxxxxxxxx> wrote:
Hi,
I'm using XSLT version 1, so can't use grouping.

I need to run through a large structure that resembles:
<sample>
   <result>
       <details>
         <group_id>250</group_id>
       </details>
   </result>
   <result>
       <details>
         <group_id>300</group_id>
       </details>
   </result>
   <result>
       <details>
         <group_id>250</group_id>
       </details>
   </result>
</sample>

Firstly, I need to sort the data into <group_id> order, so I use:
<xsl:apply-templates select="sample//result/details">
   <xsl:sort select="group_id" data-type="number" />
</xsl:apply-templates>

Next, within the called template, I need to loop through all the <details> and add row numbering information but add an extra row at the end of each group.
So hope to see:
<output row="1">250</output>
<output row="2">250</output>
<output row="3" />
<output row="4">300</output>

First problem is that, when I check "preceding::details[1]/group_id" it refers to the original document layout and not the sorted order, so I get:
<output row="1">250</output>
<output row="2" />
<output row="3">250</output>
<output row="4" />
<output row="5">300</output>

Second problem is remembering the row count, since I need to add several additional rows during the loop, so "position()" goes out of sync.
I've tried to call another template, passing in the <group_id> and determine it's sorted position but have not been able to get a unique list because of my first problem.

Hope I've made this as concise as possible.
Thanks in advance,
Carl


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--
Regards,
Mukul Gandhi

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