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AW: Beginner's question
- To: <xsl-list at lists dot mulberrytech dot com>
- Subject: AW: [xsl] Beginner's question
- From: "Michael Schafer" <MSchaefer at PFEILGMBH dot DE>
- Date: Thu, 12 Jul 2001 12:37:33 +0200
- Reply-To: xsl-list at lists dot mulberrytech dot com
Hello Chris,
try this:
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*" />
</xsl:copy>
</xsl:match>
<xsl:template match="code">
</xsl:template>
-----Ursprungliche Nachricht-----
Von: owner-xsl-list@lists.mulberrytech.com
[mailto:owner-xsl-list@lists.mulberrytech.com]Im Auftrag von J S
Publications
Gesendet: Donnerstag, 12. Juli 2001 10:25
An: XSL-List@lists.mulberrytech.com
Betreff: [xsl] Beginner's question
Hi
Here is a simple question from an absolute beginner!
Suppose I have an XML document with a complex DTD. The DTD contains a
tag
deep in the hierarchy called <code>. How would I transform the XML doc
to
another XML doc that was identical except for having the <code> tag
removed?
I said it was a simple question!
Chris
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