on 03-11-2008 1:25 PM
Hi eevrybody,
we have an XSLT that generates an XML. The generated XML should have an reference to another XSLT.
With other words: The generated XML has to look like:
<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="X:\path\path\AnotherXSLT.xslt"?> // how do do that?
<ROOT> ......
How can we do that?
Thanks Regards Mario
no replies
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Hi
Try
<xsl:processing-instruction name="xml-stylesheet">href="X:\path\path\AnotherXSLT.xslt" type="text/xml"</xsl:processing-instruction>
this should be deifned in the top level template e.g.
<xsl:template match="/">
<xsl:processing-instruction name="xml-stylesheet">href="X:\path\path\AnotherXSLT.xslt" type="text/xml"</xsl:processing-instruction>
</xsl:template>
Thanks
Damien
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