RE: [xsl] document() widlcard?

["Michael Kay" <mhkay@xxxxxxxxxxxx>]

>  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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