Hi,
You have a named template and you have the call to the template
inside the named template.
What you shoud be doing is something like:
<xsl:template match="/">
<xsl:apply-templates/>
</xsl:template>
<xsl:template match="part1">
<xsl:call-template name="part1"/>
<xsl:with-param ... etc
</xsl:template>
You may want to rename your named-template to something other than part1
though.
Cheers,
Omprakash.V
henry human
<henry_human@ To:
xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
yahoo.de> cc: (bcc: omprakash.v/Polaris)
Subject: [xsl] replacement
example , help
03/29/2005
04:31 PM
Please
respond to
xsl-list
hello,
In this xsl styesheet i will replace aa with the
string cc,
what do i wrong ,that
i dont get cc?
thank you to have a look on this stylesheet:
<?xml version="1.0"?>
<?xml-stylesheet type="text/xsl" href="replace.xsl"?>
<document>
<part1>
<name>uaabb</name>
</part1>
</document>
------------
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="html"/>
<xsl:template name="part1">
<xsl:param name="name"/>
<xsl:param name="aa"/>
<xsl:param name="cc"/>
<xsl:choose>
<xsl:when test="contains($name,$aa)">
<xsl:value-of select="substring-before($name,$aa)"/>
<xsl:value-of select="$cc"/>
<xsl:call-template name="part1">
<xsl:with-param name="name"
select="substring-after($name,$aa)"/>
<xsl:with-param name="aa" select="$aa"/>
<xsl:with-param name="cc" select="$cc"/>
</xsl:call-template>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="$name"/>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
</xsl:stylesheet>
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