Subject: [xsl] Re: Re: Retrieving XML document's name From: Dimitre Novatchev <dnovatchev@xxxxxxxxx> Date: Tue, 3 Jul 2001 01:14:20 -0700 (PDT) |
Matt Sergeant wrote: > On 03 Jul 2001 00:16:35 +0100, Sebastian Rahtz wrote: > > John Putman writes: > > > This is probably pretty simple, but I couldn't find the answer in the > > > archives or books... how can you retrieve the name of the XML file that you > > > are transforming? > > > > you can't. after all, it may not be a file at all. > > > If this was the XSLT designer's thinking on this, surely it's mistaken. > Even if it's *not* a file, you have to have given a URI to the resource, > so you should be able to retrieve that. I bet you could write an > extension function to do it. Using MSXML one can do things like: objXML.loadXML("<dynamic>" + myVar + "</dynamic>"); I have done similar things with Saxon, probably dynamically generated xml can be fed to almost any XSLT processor. Could you tell us what URI must correspond to the dynamically constructed argument of loadXML() ? Cheers, Dimitre Novatchev. __________________________________________________ Do You Yahoo!? Get personalized email addresses from Yahoo! Mail http://personal.mail.yahoo.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
Re: [xsl] Quark -> XSL -> XML, Bernard Harrison | Thread | [xsl] Re: Re: Retrieving XML docume, Matt Sergeant |
Re: [xsl] Quark -> XSL -> XML, Bernard Harrison | Date | Re: [xsl] need an "&" in my text!!!, Jeni Tennison |
Month |