Re: basic xpath question

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Subject: Re: basic xpath question
From: "Vasu Chakkera" <vasucv@xxxxxxxxxxx>
Date: Thu, 20 Jun 2002 12:25:13 +0000

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:  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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Current Thread
basic xpath question Hanlan, Dominic - Senior Developer - Thu, 20 Jun 2002 05:11:54 -0400 (EDT) Jeni Tennison - Thu, 20 Jun 2002 05:32:27 -0400 (EDT) <Possible follow-ups> Stuart Brown - Thu, 20 Jun 2002 05:35:36 -0400 (EDT) Vasu Chakkera - Thu, 20 Jun 2002 08:25:45 -0400 (EDT) <=

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