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

Subject: [xsl] Wrap changing element sequence into container: with 'for-each-group'?
From: Yves Forkl <Y.Forkl@xxxxxx>
Date: Fri, 26 Jan 2007 11:05:50 +0100

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

Current Thread
[xsl] Wrap changing element sequence into container: with 'for-each-group'? Yves Forkl - 26 Jan 2007 10:06:14 -0000 <= David Carlisle - 26 Jan 2007 10:19:36 -0000 Abel Braaksma - 26 Jan 2007 10:34:42 -0000 Yves Forkl - 30 Jan 2007 13:55:17 -0000 David Carlisle - 30 Jan 2007 14:04:10 -0000

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