Re: [xslt] how to access default xml
- From: Bernhard Zwischenbrugger <bz datenkueche com>
- To: xslt gnome org
- Subject: Re: [xslt] how to access default xml
- Date: Wed, 26 Sep 2001 14:12:45 +0200
I had some problemes with this to convert
$test-xmlfile to a node-set.
Sometimes it seems to be a result-tree-fragment.
With simple examples this works well, but I have
a strange behavior with big xml's and big xslt's.
I try to provide a more precise example.
> you want to put the original XML tree into a global variable so you can
> access it whenever you need. At the top level of the XSL, put:
>
> <xsl:variable name="default-xmlfile" select="/"/>
>
> Also, when you make repeated document function calls to the same document,
> why not put the result in a variable once and then reuse it.
>
> <xsl:variable name="test-xmlfile" select="document('test.xml')/rootnode"/>
>
> <xsl:apply-templates select="$test-xmlfile"/>
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