Subject: Re: [xsl] get the current xml file name|
From: Frank Marent <frank.marent@xxxxxxxxxxx>
Date: Tue, 8 Aug 2006 09:58:46 +0200
'/../' 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.
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
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 frank
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