[xsl] get the current xml file name

[Frank Marent <frank.marent@xxxxxxxxxxx>]

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