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Re: xsl/xml transormation

From: Stefan Wachter <Stefan.Wachter@gmx.de>
Date: Thu, 26 Sep 2002 13:25:13 +0200 (MEST)
To: abdolah@freenet.de;xmlschema-dev@w3.org;;
Message-ID: <10632.1033039513@www47.gmx.net>

I think you should post questions regarding stylesheets in other lists.
However, I think your problem is solved with the following template:

<template match="node()">
  <copy><text>
</text>
    <apply-templates/>
  </copy>
</template>

--Stefan

> 
> hello,
> I'm working with saxon to Transform xsl/xml-file to Text-Format
> In the Head of my xsl-stylesheet i used:
> <?xml version="1.0" encoding="iso-8859-1"?>
> <xsl:stylesheet version="1.0"
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
> <xsl:output method="text" indent="yes"/> 
> .....
> As Rsesult of transformation  i get:
> line1line2line3line4line5...........
> I' will but to get all breaks in my document :
> line1
> line2
> line3
> ........
> could somebody help me, how i get all line-breaks in my resultfile?
> I thank at advance
> A.A
> 
> 
> 
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Received on Thursday, 26 September 2002 07:25:48 GMT

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