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[xsl] possible? for-each-group inside a for-each-group

From: "Klocker Christoph" <Christoph.Klocker@------------->
Date: 3/2/2005 11:37:00 AM

I am trying a for-each-group inside another one, is this possible, or do I
have to do this in two following transformations??

I tried following stylesheet. Thought about adding a current-group() inside
the group-ending-with statement, didn't work either.


		<xsl:for-each-group select="*"
	<xsl:apply-templates select="current-group()"/>

My XML structure is something like:

<par class="Entsch_normal"><inline class="*fett">Sachverhalt: </inline>Die Bf
beantragte mit der Anmeldung vom 7. 9. 1999 die Ausfuhr von 21 Paletten
Fleisch von <par>
  <par class="Entsch_normal">Die Ausfuhr aus dem Zollgebiet der Gemeinschaft
erfolgte |ber das Zollamt B am 7. 9. 1999.</par>
  <par class="Entsch_normal">Mit B vom 19. 10. 1999 gewdhrte das Zollamt
S/Erstattungen der Bf eine Ausfuhrerstattung iHv 140.863 S (10.236,91
   <par class="Entsch_normal"><inline class="*fett">Erkenntnis:</inline>
<inline class="*kursiv">Abweisung</inline> als unbegr|ndet.</par>
    <par class="Entsch_normal"><inline class="*fett">Entscheidungsgr|nde:
</inline>Im Beschwerdefall erf|llten diese nicht die Voraussetzungen f|r die
Gewdhrung der differenzierten Erstattung.</par>
  <par class="Entsch_normal">Der Nachweis durch die in Art 16 Abs 1 der
genannten VO angef|hrten Urkunden zu erbringen.</par>

Where I need everything before the "Erkenntnis:"


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