Re: [xsl] xsl transform

Subject: Re: [xsl] xsl transform
From: James Fuller <jim.fuller@xxxxxxxxxxxxxx>
Date: Tue, 28 Jun 2005 16:06:19 +0200
Philippe LAPLANCHE wrote:

>Hello,
>
>Does somebody has a neet way to convert this tree :
><?xml version="1.0" encoding="UTF-8"?>
><results>
><row>
>	<param1>p1</param1>
>	<param2>p2</param2>
>	<param3>p3</param3>
>	<param4>p4</param4>
>	<param5>p5</param5>
>	<param6>p6</param6>
>	...
>      <paramn>pn</paramn>
></row>
><row>
>	...
></row>
></results>
>
>Into this one :
><?xml version="1.0" encoding="UTF-8"?>
><root fields="param1|param2|param3 ... ">
>	<e a="p1" b="p2" c="p3" d="p4" e="p5" f="p6" ... z="p26" aa="p27" ab="p28" ... az="p52" ba="p53" .... zz="p702" />
></root>
>
>
>
>  
>
something like http://www.dpawson.co.uk/xsl/sect2/N1553.html#d2005e139
will point u in the right direction.

gl, Jim Fuller

note : try to use a descriptive subject line helps those who have the
same prob later on

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