XSLT xpath problem

T

Tjerk Wolterink

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
 
D

David Carlisle

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
 
M

Mukul Gandhi

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:eek: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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