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Re: [xsl] Children with the same name


Subject: Re: [xsl] Children with the same name
From: David Carlisle <davidc@xxxxxxxxx>
Date: Mon, 21 Nov 2011 16:45:27 +0000

On 21/11/2011 16:34, Merrilees, David wrote:
Hi

I'm stuck with an Xpath. How can I select child nodes with the same name? The names are arbitrary. So far I have this, which does not work:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:output method="xml" encoding="utf-8" indent="yes"/>

   <xsl:template match="/">
     <test>
your desired output had <array> here?
<xsl:apply-templates select="array/node()[name() = name(../node())]" />
../node() selects all the children of your parent including yourself, so this would not work even if you modified it to call name(0 on each node sparately.
     </test>
   </xsl:template>

   <xsl:template match="node()">
     <xsl:copy-of select="*" />
   </xsl:template>
* copies the child elements but your examples only have child text so uou would lose the contents of each item.

</xsl:stylesheet>


Input

<array>
   <item/>
your desired result does not have this empty item even though it shares the name "item", is there an additioanl unstated rule that empty elements be filtered, or should this be in the output?

   <nope/>
   <item>4</item>
   <item>5</item>
   <wibble/>
   <item>six</item>
   <wibble>stuff</wibble>
   <item>item</item>
   <no/>
</array>


as with all grouping, this would be easier in xslt2, but assuming you do need 1, something like


<xsl:key name="n" match="*" use="concat(generate-id(..),name())"/>

select="array/*[key('n',concat(generate-id(..),name())[2]]" />

David

Desired output

<array>
   <item>4</item>
   <item>5</item>
   <wibble/>
   <item>six</item>
   <wibble>stuff</wibble>
   <item>item</item>
</array>


Thanks


-----Original Message-----
From: Merrilees, David
Sent: 18 October 2011 16:46
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: RE: [xsl] required parameter

It returns

<form id="fLanguage" action="/en-GB/Language/Set" method="post"></form>

Thanks

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