Subject: Re: [xsl] For-each-group problem From: "Mukul Gandhi" <gandhi.mukul@xxxxxxxxx> Date: Mon, 2 Oct 2006 23:03:34 +0530 |
<root> <body> <heading/> <p></p> <p></p> <li></li> <div> <p></p> <p></p> </div> <li></li> <div> <p></p> <p></p> <p></p> <p></p> </div> <li></li> <div> <p></p> <p></p> </div> <p></p> <p></p> <footer></footer> </body> </root>
<?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="node() | @*"> <xsl:copy> <xsl:apply-templates select="node() | @*" /> </xsl:copy> </xsl:template>
<xsl:template match="div"> <li> <xsl:copy-of select="." /> </li> </xsl:template>
As Andrew said, you don't need grouping here, but a variation of identity transform.
Can't seem to get the results I want using for-each-group
Using Saxon 8 and XSLT 2
Sample doc
<root> <body> <heading> <p></p> <p></p>
<li></li> <div> <p></p> <p></p> </div>
<li></li> <div> <p></p> <p></p> <p></p> <p></p> </div>
<li></li> <div> <p></p> <p></p> </div>
<p></p> <p></p> <footer></footer> <body> </root>
I want it to look like the following
<root> <body> <heading> <p></p> <p></p>
<li> <div> <p></p> <p></p> </div> </li>
<li> <div> <p></p> <p></p> <p></p> <p></p> </div> </li>
<li> <div> <p></p> <p></p> </div> </li>
<p></p> <p></p> <footer></footer> <body> </root>
I've followed the example in MK's XSLT 2 book page # 297 and haven't achieved the same results. I have a bunch of elements that this must be done to.
Any help would be grealy appreciated.
Mario Madunic
-- Regards, Mukul Gandhi
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