hi michael now works fine.
thank you all
--- Michael Kay <mike(_at_)saxonica(_dot_)com> schrieb:
You misunderstood the message and have integrated
the new code incorrectly.
You need your original named template, which does
the string replacement,
and you need code that invokes it when you get to a
suitable place in the
source document. You also need to supply values for
the parameters. So:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="html"/>
<xsl:template match="/">
<xsl:apply-templates/>
</xsl:template>
<xsl:template match="document">
<xsl:copy><xsl:apply-templates/></xsl:copy>
</xsl:template>
<xsl:template match="part1">
<xsl:call-template name="part1">
<xsl:with-param name="name" select="name"/>
<xsl:with-param name="aa" select="'aa'"/>
<xsl:with-param name="cc" select="'cc'"/>
</xsl:call-template>
</xsl:template>
<xsl:template name="part1">
... your original code...
</xsl:template>
</xsl:stylesheet>
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: henry human [mailto:henry_human(_at_)yahoo(_dot_)de]
Sent: 29 March 2005 13:37
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] replacement example , help
i still becomm error in IE: part1 template has'nt
exist.
??
here new code as you said:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="html"/>
<xsl:template match="/">
<xsl:apply-templates select="part1"/>
</xsl:template>
<xsl:template match="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>
--- omprakash(_dot_)v(_at_)polaris(_dot_)co(_dot_)in schrieb:
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"?>
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