Re: [xsl] basic xpath question

["Vasu Chakkera" <vasucv@xxxxxxxxxxx>]

Try this code..
Hope this helps..
<?xml version="1.0"?>
<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output omit-xml-declaration="yes" media-type="html"/>
<xsl:template match="/">
<p>
<xsl:for-each select ="//section">
<xsl:choose>
<xsl:when test="./@id = 'url'">
<xsl:element name="a">
<xsl:attribute name="href"><xsl:value-of select="./@url"/></xsl:attribute>
<xsl:value-of select="."/></xsl:element>
</xsl:when>
<xsl:otherwise><xsl:value-of select="text()"/></xsl:otherwise>
</xsl:choose>
</xsl:for-each>
</p>
</xsl:template>
</xsl:stylesheet>

Cheers
Vasu

> From: "Hanlan, Dominic - Senior Developer" <dhanlan@xxxxxxxxx>
> Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> To: "'XSL-List@xxxxxxxxxxxxxxxxxxxxxx'" <XSL-List@xxxxxxxxxxxxxxxxxxxxxx>
> Subject: [xsl] basic xpath question
> Date: Thu, 20 Jun 2002 10:11:25 +0100
>
> Hi,
>
>  I have an xml document to be styled, if I have a construct such as
>
> <section id="para">this is text <section id="url"
> url="http://xx.yy.zz";>which is to be a link</section></section>
>
> How do I, in xslt, recognise the "url" tag within the "para" tag, such that
> I could ouput someting like
>
> <p>this is test <a href="http://xx.yy.zz";>which is to be a link</a></p>
>
> Regards
>
>
>
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list




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