Hi Batis,
You cannot use call-template with select="/root/bib"
"Unlike xsl:apply-templates, xsl:call-template does not change the
current node or the current node list." [1]
You should give a try to using apply-templates instead.
Besides, "variable" in XSLT cannot be changed after first initialization, so
<xsl:variable name="var" select="concat($var,ids, $delimiter)"/> is
bound to fail.
[1] XSL Transformation (XSLT) version 1.0, §6 Named Templates
http://www.w3.org/TR/xslt#named-templates
Cheers,
Eric Bréchemier
On Jan 8, 2008 4:58 PM, Batis DAVE <batis(_dot_)04(_at_)gmail(_dot_)com> wrote:
I want to return a variable value from a template based on this example:
http://biglist.com/lists/xsl-list/archives/200205/msg01614.html
but it's not working...
Here's my code:
<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">
<xsl:output indent="yes" method="html" omit-xml-declaration="no"
encoding="ISO-8859-1" />
<xsl:param name="delimiter" select="','"/>
<xsl:template match="/root">
<xsl:variable name="idList">
<xsl:call-template select="/root/bib"
name="getIdList" />
</xsl:variable>
<xsl:value-of select="$idList"/>
</xsl:template>
<xsl:template match="/root/bib" name="getIdList">
<xsl:for-each select="/root/bib">
<xsl:variable name="var" select="concat($var,ids,
$delimiter)"/>
</xsl:for-each>
<xsl:value-of select="$var"/>
</xsl:template>
</xsl:stylesheet>
What's wrong in it?
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