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Hi,
Tempore 14:02:26, die 07/05/2005 AD, hinc in xsl-list@xxxxxxxxxxxxxxxxxxxxxx scripsit Lakshmi narayana <lchintala@xxxxxxxxxxxx>:
try some stylesheet like this:
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="node() | @*">
<xsl:copy>
<xsl:attribute name="id"><xsl:number count="*" level="any"/></xsl:attribute>
<xsl:apply-templates select="node() | @*" />
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
Re: [xsl] Generating Sequential IDs
Subject: Re: [xsl] Generating Sequential IDs From: "Joris Gillis" <roac@xxxxxxxxxx> Date: Tue, 05 Jul 2005 13:40:15 +0200 |
Hi,
Tempore 14:02:26, die 07/05/2005 AD, hinc in xsl-list@xxxxxxxxxxxxxxxxxxxxxx scripsit Lakshmi narayana <lchintala@xxxxxxxxxxxx>:
I want to generate the sequence ids for all the nodes in the xml tree.
try some stylesheet like this:
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="node() | @*">
<xsl:copy>
<xsl:attribute name="id"><xsl:number count="*" level="any"/></xsl:attribute>
<xsl:apply-templates select="node() | @*" />
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
regards, -- Joris Gillis (http://users.telenet.be/root-jg/me.html) Spread the wiki (http://www.wikipedia.org)
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