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RE: [xsl] document() widlcard?
Subject: RE: [xsl] document() widlcard? From: "Michael Kay" <mhkay@xxxxxxxxxxxx> Date: Thu, 9 Aug 2001 09:32:56 +0100 |
> 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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