XSLT xpath problem

Tjerk Wolterink · Jul 14, 2005

I have xml like this:

<data>
<item><type>a</type></item>
<item><type>a</type></item>
<item><type>b</type></item>
<item><type>b</type></item>
<item><type>b</type></item>
<item><type>c</type></item>
</data>

I want an xsl document that transforms this in this:

<ul>
<li>a</li>
<li>b</li>
<li>c</li>
</ul>

My xsl:

<xsl:template match="data">
<uL>
<xsl:for-each select="item">
<xsl:if test=" the type of the item before this item is different">
<li><xsl:value-of select="type"/></li>
</xsl:if>
</xsl:for-each>
</ul>
</xs:template>

pleaze help

Dimitre Novatchev · Jul 14, 2005

This is a grouping problem. For more information see:

http://jenitennison.com/xslt/grouping/index.html

Cheers,
Dimitre Novatchev.

David Carlisle · Jul 14, 2005

A Dimitre says this is a grouping problem but if you are in the special
case that you know your input is already in the right order, you can
code it more or less exactly as you sketched, without having to use
grouping at the xslt level

<xsl:if test=" the type of the item before this item is different">

is

<xsl:if test="not(type = preceding-sibling::item[1]/type)">

David

Mukul Gandhi · Jul 14, 2005

You may try this. This is Muenchian grouping, and is efficient. Or you
may use David's method.

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

<xsl

utput method="html" indent="yes" />

<xsl:key name="by-type" match="item/type" use="." />

<xsl:template match="/data">
<html>
<head>
<title/>
</head>
<body>
<ul>
<xsl:for-each select="item/type[generate-id() =
generate-id(key('by-type', .)[1])]">
<li><xsl:value-of select="." /></li>
</xsl:for-each>
</ul>
</body>
</html>
</xsl:template>

</xsl:stylesheet>

Regards,
Mukul

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XSLT xpath problem

Tjerk Wolterink

Dimitre Novatchev

David Carlisle

Mukul Gandhi

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