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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