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Postnext
Jeff HatchSubject: List item grouping in XSLT 2.0
Author: Jeff Hatch
Date: 03 Jun 2005 12:48 PM
Hello,

I have tried to get my arms around the new grouping features in XSLT 2.0, but am having some difficulty. Need to transform the following:

<chapter>
<head1>This is the Head 1</head1>
<para>This is paragraph content</para>
<para>This is paragraph content</para>
<para>This is paragraph content</para>
<head1>This is the head 1</head1>
<bulletListItem>• This is a bullet list item</bulletListItem>
<bulletListItem>• This is a bullet list item</bulletListItem>
<bulletListItem>• This is a bullet list item</bulletListItem>
<bulletListItem>• This is a bullet list item</bulletListItem>
<para>This is paragraph content</para>
<para>This is paragraph content</para>
<bulletListItem>• This is a bullet list item</bulletListItem>
<bulletListItem>• This is a bullet list item</bulletListItem>
<Head2>This is the head 2</Head2>
<numberListItem>1. This is a number list item</numberListItem>
<numberListItem>2. This is a number list item</numberListItem>
<numberListItem>3. This is a number list item</numberListItem>
<numberListItem>4. This is a number list item</numberListItem>
<numberListItem>5. This is a number list item</numberListItem>
</chapter>

To the following output/result:

<chapter>
<head1>This is the Head 1</head1>
<para>This is paragraph content</para>
<para>This is paragraph content</para>
<para>This is paragraph content</para>
<head1>This is the head 1</head1>
<bulletList>
<bulletListItem>• This is a bullet list item</bulletListItem>
<bulletListItem>• This is a bullet list item</bulletListItem>
<bulletListItem>• This is a bullet list item</bulletListItem>
<bulletListItem>• This is a bullet list item</bulletListItem>
</bulletList>
<para>This is paragraph content</para>
<para>This is paragraph content</para>
<bulletList>
<bulletListItem>• This is a bullet list item</bulletListItem>
<bulletListItem>• This is a bullet list item</bulletListItem>
</bulletList>
<Head2>This is the head 2</Head2>
<numberList>
<numberListItem>1. This is a number list item</numberListItem>
<numberListItem>2. This is a number list item</numberListItem>
<numberListItem>3. This is a number list item</numberListItem>
<numberListItem>4. This is a number list item</numberListItem>
<numberListItem>5. This is a number list item</numberListItem>
</numberList>
</chapter>

Any help would be greatly appreciated. Thanks ...

Postnext
Ivan PedruzziSubject: List item grouping in XSLT 2.0
Author: Ivan Pedruzzi
Date: 06 Jun 2005 08:15 AM
Hi Jeff,

The following method works for both XSLT versions

Hope this helps
Ivan Pedruzzi
Stylus Studio Team

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:template match="/">
<xsl:apply-templates/>
</xsl:template>

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

<xsl:template match="*" mode="body">
<xsl:copy-of select="."/>
<xsl:variable name="local" select="local-name()"/>
<xsl:apply-templates select="following-sibling::*[1][local-name() = $local]" mode="body"/>
</xsl:template>

<xsl:template match="bulletListItem">
<xsl:if test="local-name(preceding-sibling::*[1]) != 'bulletListItem'">
<bulletList>
<xsl:apply-templates select="." mode="body"/>
</bulletList>
</xsl:if>
</xsl:template>

<xsl:template match="numberListItem">
<xsl:if test="local-name(preceding-sibling::*[1]) != 'numberListItem'">
<numberList>
<xsl:apply-templates select="." mode="body"/>
</numberList>
</xsl:if>
</xsl:template>

</xsl:stylesheet>

Postnext
Jeff HatchSubject: List item grouping in XSLT 2.0
Author: Jeff Hatch
Date: 06 Jun 2005 10:43 AM
This was very helpful -- thank you. I'm still curious to know if this sort of grouping could be handled with the "for-each-group" process in 2.0. My attempts have failed, in that if I have two separate bulleted lists, my stylesheet grouped all list items together in the same single list. Anyone have any thoughts?

Posttop
Ivan PedruzziSubject: List item grouping in XSLT 2.0
Author: Ivan Pedruzzi
Date: 08 Jun 2005 01:57 AM

Here is a solution using for-each-group (XSLT 2.0)

Hope this helps
Ivan Pedruzzi
Stylus Studio Team


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

<xsl:template match="/">
<xsl:apply-templates/>
</xsl:template>

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

<xsl:template match="chapter">
<xsl:copy>
<xsl:for-each-group select="*" group-adjacent="local-name()">
<xsl:choose>
<xsl:when test="current-grouping-key() = 'bulletListItem'">
<bulletList>
<xsl:copy-of select="current-group()"/>
</bulletList>
</xsl:when>
<xsl:when test="current-grouping-key() = 'numberListItem'">
<numberList>
<xsl:copy-of select="current-group()"/>
</numberList>
</xsl:when>
<xsl:otherwise>
<xsl:apply-templates select="current-group()"/>
</xsl:otherwise>
</xsl:choose>
</xsl:for-each-group>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>

   
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