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| | | Altova Mailing List Archives>Archive Index >xsl-list Archive Home >Recent entries >Thread Prev - RE: [xsl] Different results of format-date() when using empty seq. or omitting args for lang/cal/country args [Thread Next] Re: [xsl] For-each-group problemTo: Date: 10/2/2006 5:34:00 PM Your XML is not well formed. But assuming that your well formed XML would be:
<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>
The following stylesheet produces the wanted output:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes" />
<xsl:template match="node() | @*">
<xsl:copy>
<xsl:apply-templates select="node() | @*" />
</xsl:copy>
</xsl:template>
<xsl:template match="li" />
<xsl:template match="div">
<li>
<xsl:copy-of select="." />
</li>
</xsl:template>
</xsl:stylesheet>
As Andrew said, you don't need grouping here, but a variation of
identity transform.
On 10/2/06, Mario Madunic <hajduk@xxxxxxxx> wrote:
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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