Re: [xsl] Template matching similiar names

["Vasu Chakkera" <vasucv@xxxxxxxxxxx>]

This should be easy i think...
Tru this one

<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:template match="/">
<xsl:apply-templates select="classes/class[@element = 'root']"/>
</xsl:template>
<xsl:template match="class">
<class element = "{@element}">
<doc>Docs for root</doc>
</class>
</xsl:template>
</xsl:stylesheet>
note that i have used a parent element for class elements. I have assumed
the xml structure as follows..
<?xml version="1.0"?>
<classes>
<class element="root">
</class>

<class element="roothome">
</class>
</classes>
Do proper alterations to reflect your xml and it will work
cheers
HTH
Vasu Chakkera
----- Original Message -----
From: "Holbrook, R Cody (Cody)" <rch7@xxxxxxxxx>
To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
Sent: Monday, April 07, 2003 6:04 PM
Subject: [xsl] Template matching similiar names


> Hello All,
>
> I'm having difficulty distinguishing between classes with similiar names.
>
> I start with the following XML
>
> <class element="root">
> </class>
>
> <class element="roothome">
> </class>
>
> The template match I'm using is incorrect as I only want to have the doc
tags added to the class with the element attribute root.
>
>
> What I get back is this:
>
> <class element="root">
> <doc>Docs for root</doc>
> </class>
>
> <class element="roothome">
> <doc>Docs for root</doc>
> </class>
>
>
> No matter what I've done, I get the same thing, in general my tries have
looked like this:
>
> <xsl:template match="xsd:class...">
> <xsl:copy>
> <xsl:copy-of select="@*"/>
> <doc>Docs for root</doc>
> ....
>
> Is there a way to make a very exact match with an attribute?
>
> Thanks,
>
> Cody Holbrook
>
>
>
> (and many variations, including two starts-with template matches that
worked, but not well)
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
>
>

 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list