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copying <xsl:stylesheet> tag to output xsl file
- From: "tony" <dhirajtorane at yahoo dot com>
- To: <xsl-list at lists dot mulberrytech dot com>
- Date: Wed, 2 Jan 2002 19:39:59 +0530
- Subject: [xsl] copying <xsl:stylesheet> tag to output xsl file
- Reply-to: xsl-list at lists dot mulberrytech dot com
Hi,
I have this XSL file as below..i want to copy the <xsl:stylesheet> tag to
the output xsl file, as i need the "my" namespace declaration in the
generated output xsl file.
<xsl:stylesheet version="1.0" xmlns:my="http://mysite.com/mynamespace">
<xsl:output method="xml" version="1.0" indent="yes"/>
<xsl:template match="/">
<xsl:for-each select="document('')/xsl:stylesheet">
<xsl:copy>
<xsl:apply-templates/>
</xsl:copy>
</xsl:for-each>
</xsl:template>
When i apply this XSL on a XML , it copies the <xsl:stylesheet> element
properly, but the <xsl:apply-template> applies the template on to this
same XSL file instead of the XML file.
e.g. The templates are applied on to <xsl:output> , <xsl:template>,
<xsl:for-each>, <xsl:copy> tags which are actually in the XSL file.
Can some one tell me a way by which i can copy the <xsl:stylesheet> element
to the output XSL file, & apply the templates on to the XML file so that the
output of comes between
<xsl:stylesheet version="1.0" xmlns:my="http://mysite.com/mynamespace">
<!-- output from the XML file after applying the templates -->
</xsl:stylesheet>
tags.
I am sorry but i am not able to form the question properly.
Thanks,
Dhiraj
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