Thanks to everyone for your solutions.

I tried them all but got the problem of repeating DOG and LION elements
in my output.  Couldn't figure it out why.

The only solution that truly worked perfectly for me was the one from
Martin.
Many Thanks again!


Houman


-----Original Message-----
From: Martin Honnen [mailto:Martin.Honnen@xxxxxx]
Sent: Monday, October 01, 2007 2:21 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re: [xsl] Reading recursive from a list

Houman Khorasani wrote:

> My Selection list:
>
> <UnitInfos xmlns="Animals">
> 	<Type>UNIT_LION</Type>
> 	<Type>UNIT_DOG</Type>
> </UnitInfos>
>
>
> According to my selection List, I would like to copy all UnitInfo
> elements with their children that have a type element which is listed
in
> the small list above.


Here is an XSLT 1.0 stylesheet that has the "selection list" in the
stylesheet:

<xsl:stylesheet
   xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
   xmlns:a="Animals"
   exclude-result-prefixes="a"
   version="1.0">

   <xsl:output method="xml" indent="yes"/>
   <xsl:strip-space elements="*"/>

   <UnitInfos xmlns="Animals">
	<Type>UNIT_LION</Type>
	<Type>UNIT_DOG</Type>
   </UnitInfos>

   <xsl:template match="@* | node()">
     <xsl:copy>
       <xsl:apply-templates select="@* | node()"/>
     </xsl:copy>
   </xsl:template>

   <xsl:template match="a:UnitInfos">
     <xsl:copy>
       <xsl:apply-templates select="@*"/>
       <xsl:apply-templates select="a:UnitInfo[a:Type =
document('')/xsl:stylesheet/a:UnitInfos/a:Type]"/>
     </xsl:copy>
   </xsl:template>

</xsl:stylesheet>


--

	Martin Honnen
	http://JavaScript.FAQTs.com/


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