Re: [xsl] Wrap changing element sequence into container: with 'for-each-group'?

[Abel Braaksma <abel.online@xxxxxxxxx>]

Hi Yves,

You did not show how the output should really look like. But I suppose 
"inside a <b> container" means a <b> element with children <b1>, <b2> 
etc. Here's what I came up with, which does not involve grouping, but 
instead uses mode-switching. If you have textual content in your real 
input, you may need some changes, because I neglect the default template 
below.

Output of below stylesheet on your input:

<a>
  <b>
     <b1/>
     <b2/>
  </b>
  <c/>
</a>
<a>
  <b>
     <b2/>
  </b>
  <c/>
</a>


The Stylesheet:

<xsl:stylesheet
   xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
   version="2.0">

   <xsl:output indent="yes" />
  
   <xsl:template match="/">
       <xsl:apply-templates select="$abc/*" />
   </xsl:template>
  
   <xsl:template match="*[starts-with(name(), 'b')][1]">
       <b>
           <xsl:apply-templates
               select="self::* | following-sibling::*"
               mode="b"/>
       </b>
   </xsl:template>
  
   <xsl:template match="*[starts-with(name(), 'b')]" mode="b">
       <xsl:copy>
           <xsl:apply-templates select="node() | @*" mode="#current"/>
       </xsl:copy>
   </xsl:template>

   <!-- copy all others -->
   <xsl:template match="*[not(starts-with(name(), 'b'))]">
       <xsl:copy>
           <xsl:apply-templates select="node() | @*" />
       </xsl:copy>
   </xsl:template>
  
</xsl:stylesheet>



Cheers,
-- Abel Braaksma
  http://www.nuntia.nl

Yves Forkl wrote:
> Given two XML fragments
>
> <a>
>  <b1/>
>  <b2/>
>  <c/>
> </a>
>
> and
>
> <a>
>  <b2/>
>  <c/>
> </a>
>
> I would like to wrap <b1/><b2/> into a <b/> container, and the same 
> should happen to <b2/> when occurring alone. The other elements can be 
> assumed to remain unchanged.
>
> While I found quite some hints on solving grouping problems in the 
> XSLT FAQ and in the XSL list archive, I can't figure out how to solve 
> this wrapping task using XSLT 2.0. There is an interesting posting 
> here: http://www.biglist.com/lists/xsl-list/archives/200201/msg00804.html
> But I don't know how this could serve as a basis for a solution to my 
> problem.
>
> I suspect that "for-each-group" should be my friend; how would I need 
> to use it here? Or is there any other, more elegant solution?
>
> Yours,
>
>   Yves