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RE: document() widlcard?
- To: <xsl-list at lists dot mulberrytech dot com>
- Subject: RE: [xsl] document() widlcard?
- From: "Michael Kay" <mhkay at iclway dot co dot uk>
- Date: Thu, 9 Aug 2001 09:32:56 +0100
- Reply-To: xsl-list at lists dot mulberrytech dot com
> 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
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