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Re: position within same nodes
- To: <xsl-list at lists dot mulberrytech dot com>
- Subject: Re: [xsl] position within same nodes
- From: "Thomas B. Passin" <tpassin at mitretek dot org>
- Date: Thu, 19 Jul 2001 15:52:42 -0400
- References: <Pine.HPX.4.33.0107191330560.3598-100000@ux1>
- Reply-To: xsl-list at lists dot mulberrytech dot com
Stylesheet:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<root>
<xsl:variable name='meeps' select='//meep'/>
<xsl:for-each select='$meeps'>
<xsl:value-of select='position()'/>--
</xsl:for-each>
</root>
</xsl:template>
</xsl:stylesheet>
Results:
<root>1--
2--
3--
</root>
Cheers,
Tom P
[Laurence O Garfield]
> What is the XPath to get the position of a given node relative to its
> siblings of the same name? Eg,
>
> <blah>
> <argh/>
> <meep/>
> <meep/>
> <stuff/>
> <meep/>
> </blah>
>
> I'm looking for an XPath statement that would return 1 for the first meep,
> 2 for the second, and 3 for the 3rd meep, rather than 2, 3, and 5 (which
> is what plain position() returns).
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