Re: [xsl] get the current xml file name

[Frank Marent <frank.marent@xxxxxxxxxxx>]

colin

you saved my day. that works! with a small change

  document(concat(document-uri(/),'/../','nextfile.xml'))

'/../' instead of '../' and a little bit wondering why there is no
function available in xslt to get the current xml file name. thanks
also to abel for his contribution... no i can't use parameters in
this case, i forgot to mention.

frank


Am 08.08.2006 um 08:42 schrieb Colin Adams:

> I would think something like:
>
> document (concat(document-uri(/),'../','nextfile.xml')) should do
> the trick.
>
> If you are using XSLT 2.0, you can use the resolve-uri function
> which will be a bit neater.
>
> > From: Frank Marent <frank.marent@xxxxxxxxxxx>
> > Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> > Subject: [xsl] get the current xml file name
> > Date: Tue, 8 Aug 2006 08:18:37 +0200
> >
> > hi
> >
> > this is my first (xslt-beginner) question to this list, hoping my
> > question is not been answered 1 mio times. i was searching in the
> > archive to avoid this. :-)
> >
> > i'm looking for a function that helps me to read and process
> > several .xml files by document(). the problem is that when i say
> >
> >   document('nextfile.xml')
> >
> > the document() function tries to read the xml file in the path of
> > the  current xslt (!) file. and that's not working for me. so i
> > want to  redirect the document() function to the path of the
> > directory of the  current .xml file.
> >
> > how can i do that?
> >
> >   document-uri(/)
> >
> > reports the whole (!) uri of the current xml file. that includes
> > the  path i'm looking for, like
> >
> >   file:/C:/somewhere/anywhere/whereami/myfile.xml
> >
> > and that's fine. but how can i now cut off 'myfile.xml' when
> > 'myfile.xml' is always changing and i have no idea of this
> > filename  in the xslt and replace it with 'nexfile.xml'?
> >
> > i thought to find a function to
> >
> >   get the current xml file name
> >
> > to be able to extract that file from the uri-path-string. but i
> > could  not find anything about this. also trying to find a
> > function to find  the last '/' in a string and extracting from
> > that with substring-after (). uahh.
> >
> > i hope to find help here. any hint is very appreciated!
> >
> > from switzerland
> > frank
>
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