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document() and position()
- To: <xsl-list at mulberrytech dot com>
- Subject: document() and position()
- From: "Dirk Holstein" <dirk dot holstein at doubleSlash dot de>
- Date: Fri, 21 Jul 2000 21:46:43 +0200
- Reply-To: xsl-list at mulberrytech dot com
I have multiple xml files. The XML Stream I get
gives me the path to the xml files and the name
of each xml file.
The output I generate doesn't need any of the xml file.
An attribute in the root tag of every xml file
decide if I need this file or not.
The access to the xml files is something like this:
<xsl:apply-templates select="document(/page/folder)/news[@show = 'true']" />
With this call I get only the xml files which have
the 'show' attribute in the 'news' tag set to 'true'.
But now I need to know the number of the actual file I
read in the template I call. I tried something like this:
<xsl:apply-templates select="document(/page/folder)/news[@show = 'true']">
<xsl:with-param name="index"><xsl:value-of select="position()"
/></xsl:with-param>
</xsl:apply-templates>
But the index parameter is allways '1'. How can I get
the position? Or is there any other way to get the
number of the actual xml file?
Thanks in advance...
Dirk Holstein
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