Re: [xsl] get the current xml file name

Subject: Re: [xsl] get the current xml file name
From: Abel Online <>
Date: Tue, 08 Aug 2006 08:54:18 +0200
Perhaps you can use xslt parameters and pass to xslt the absolute filename to your source documents with these parameters? Then you are not stuck with any path.

Abel Braaksma

Frank Marent wrote:


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


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?


reports the whole (!) uri of the current xml file. that includes the path i'm looking for, like


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

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