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Re: xml include
- To: xsl-list at lists dot mulberrytech dot com
- Subject: Re: [xsl] xml include
- From: David Carlisle <davidc at nag dot co dot uk>
- Date: Wed, 11 Apr 2001 12:31:22 +0100
- References: <002901c086c0$0f143800$3f673c3e@mhkhome> <011601c086c3$9f9dd680$da1539d4@prepro.turk.net> <200101251218.MAA20616@nag.co.uk> <000901c086d3$6c72acd0$da1539d4@prepro.turk.net> <200101251400.OAA24922@nag.co.uk> <006a01c0c276$e53243e0$da1539d4@prepro.turk.net>
- Reply-To: xsl-list at lists dot mulberrytech dot com
Hi,
I have two xml files, one of them is included in other:
<?xml version="1.0"?>
<document xmlns:xinclude="http://www.w3.org/1999/XML/xinclude">
<xinclude:include href="xml.xml"/>
</document>
And I use this xsl to transform:
<?xml version="1.0"?>
<xsl:stylesheet version='1.0'
xmlns:xsl='http://www.w3.org/1999/XSL/Transform' >
<xsl:output method="html" />
<xsl:template match="/">
<H1><xsl:value-of select="//title"/></H1>
<H2><xsl:value-of select="//author"/></H2>
</xsl:template>
</xsl:stylesheet>
(xml.xml contains title and author tags)
The problem is that xsl cannot read title and author tags ... ?
The input to XSL is (logically) a tree. If your XML parser natively
supported xml:include (do any?) and "expanded includes" the way that
it expands entities, then the tree would have <author> as a descendent
of the root node of the input document and your stylesheet would work.
As it is, xml:include is probably just seen as a general element by all
process concerned so you have to follow the link yourself.
<?xml version="1.0"?>
<xsl:stylesheet version='1.0'
xmlns:xsl='http://www.w3.org/1999/XSL/Transform'
xmlns:xinclude="http://www.w3.org/1999/XML/xinclude">
<xsl:output method="html" />
<xsl:template match="/">
<xsl:variable name="x" select="document(document/xinclude:include/@href)"/>
<H1><xsl:value-of select="$x//title"/></H1>
<H2><xsl:value-of select="$x//author"/></H2>
</xsl:template>
</xsl:stylesheet> >
David
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