Subject: RE: document() widlcard?
From: "Michael Kay" <mhkay@xxxxxxxxxxxx>
Date: Thu, 9 Aug 2001 09:32:56 +0100
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> Is there anyway to apply the following template to all of my
> xmls without
> having to reference the xsl within each? I would like to
> provide a wildcard
> within document such as document('xml/*.xml). Is there a way
> to do this?
No, there's no direct way of doing this. Your best bet is probably to write
a little app (or shell script) which constructs an XML document containing
the list of documents you want processed, e.g.
<dir>
<doc>doc1.xml</doc>
<doc>doc2.xml</doc>
<doc>doc3.xml</doc>
</dir>
And then in your stylesheet use:
select="document(document('dir.xml')/dir/doc)"
If you're feeling smart (and using a JAXP-conformant processor) you could
write a URIResolver that constructs the dir.xml document dynamically on
request.
Mike Kay
Software AG
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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Michael Kay - Thu, 9 Aug 2001 04:32:22 -0400 (EDT) <=
Hewko, Doug - Thu, 9 Aug 2001 09:38:34 -0400 (EDT)
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