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On 9/1/06, Frank Marent <frank.marent@xxxxxxxxxxx> wrote: i'm struggling again and again over this point: is there a function in xslt that gives me the clean filename of the processed xml file? like 'myfile.xml' 'test.xml' 'anyfilename.xml' i do *not* need the document-uri. only the filename of the current processed xml file. In xslt 2.0 I use: tokenize(base-uri(.), '/')[last()] In xslt 1.0 you'll need to pass it in as a parameter.

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Re: [xsl] function for getting filename