what about
<my-include uri="page1.xml">
<xinclude:include xinclude:parse="xml" xinclude:href="page1.xml"/>
</my-include>
But if the files are big, wouldn't it be better to use
document(page1.xml,title) to get exactly the titles?
Your index could also look like:
<page>
<item url="pageN.xml"/>
...
</page>
and you could load what you want using document({@url},title) in a template
matching item
Olivier
P.S. pas trop chaud à Sophia-Antipolis ;) ?
> -----Message d'origine-----
> De : Ludovic Maitre [mailto:[EMAIL PROTECTED]]
> Envoyé : lundi 30 juillet 2001 19:56
> À : Cocoon Users
> Objet : xinclude before xslt, how retrieve the url ?
>
>
> Hello Cocoon users,
>
> I want to include some files in a xml file in Cocoon 1.8.2.
> For example this index.xml :
>
> <?xml version="1.0"?>
> <?xml-stylesheet href="index.xsl" type="text/xsl"?>
>
> <?cocoon-process type="xinclude"?>
> <?cocoon-process type="xslt"?>
>
> <page xmlns:xinclude="http://www.w3.org/1999/XML/xinclude"; >
> <xinclude:include xinclude:parse="xml" xinclude:href="page1.xml"/>
> <xinclude:include xinclude:parse="xml" xinclude:href="page2.xml"/>
> <xinclude:include xinclude:parse="xml" xinclude:href="page3.xml"/>
> </page>
>
> and i want to apply a xsl who transform all these includes in
> a summary,
> a list of all the title elements contained in the included XML pages
> with
> an hyperlink to these pages.
>
> Anybody know how i can retrieve the url of the included files from
> index.xsl
> for generate a <a href> tag ?
> [i have test with the document() function of XSLT, for obtaining the
> URI-basename
> of the nodes that i traverse but i haven't succedeed to obtain the url
> of the nodes]
>
> Obviously, the purpose is to make a menu, perhaps it's not
> the good way
> if this is the case i appreciate any suggestion.
>
> Thanks per advance and good evening (or day),
>
> --
> [EMAIL PROTECTED]
>
> INRIA - 2004 route des lucioles - BP 93 Tel: (33/0) 4 92 38 50 41
> 06902 SOPHIA-ANTIPOLIS cedex (France) Fax: (33/0) 4 92 38 76 02
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