[xsl] get the current xml file name

Subject: [xsl] get the current xml file name
From: Frank Marent <frank.marent@xxxxxxxxxxx>
Date: Tue, 8 Aug 2006 08:18:37 +0200

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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