Re: [xsl] sorting AND copying of XML via XSL

Subject: Re: [xsl] sorting AND copying of XML via XSL
From: Mukul Gandhi <mukulw3@xxxxxxxxx>
Date: Tue, 9 Sep 2003 10:48:39 -0700 (PDT)
Hi Marcel,
  Below is the complete XSL --

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="xml" version="1.0"
encoding="UTF-8" indent="yes"/>

<xsl:template match="/foo">
  <foo>
    <xsl:for-each select="bar">
     <xsl:sort select="@id" data-type="number"/>
	<bar id="{@id}">
	  <xsl:for-each select="item">
	   <xsl:sort select="@id" data-type="number"/>
		<item id="{@id}">
		  <xsl:value-of select="."/>
		</item>
	  </xsl:for-each>
	</bar>
    </xsl:for-each>
  </foo>
</xsl:template>
</xsl:stylesheet>

Regards,
Mukul


--- marcel salathe <ms7141@xxxxxx> wrote:
> Hi
> 
> I have a XSL question which might be easy to answer,
> but I couldnt.
> 
> Suppose I have an XML document like this:
> 
> <foo>
>   <bar id="2">
>     <item id="6">hello</item>
>     <item id="4">world</item>
>   </bar>
>   <bar id="1">
>     <item id="7">how's</item>
>     <item id="3">life</item>
>   </bar>
> </foo>
> 
> Now I want to sort the bar elements and then the
> item elements according to
> their Id's, and finally output 
> the entire document sorted, e.g.
> 
> <foo>
>   <bar id="1">
>     <item id="3">life</item>
>     <item id="7">how's</item>
>   </bar>
>   <bar id="2">
>     <item id="4">world</item>
>     <item id="6">hello</item>
>   </bar>
> </foo>
> 
> I could achieve sorting or copying, but I couldnt do
> both together. I guess
> it must be quite a simple 
> XSLT. 
> 
> Any help very much appreciated.
> 
> Best regards,
> marcel
> 
> 
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