xslt - get file name of xml file with xsl -


what's best way name of accessed xml document xsl 2.0?

i guess it's combination of resolve-uri , base-uri.

base-uri gives me absolute path, need name of file, without path. there smart way wiithout substring-before , stuff that?

so when path c:/users/abc/desktop/somefile.xml, need somefile.xml.

thanks , tips!

how about:

tokenize(base-uri(), '/')[last()]

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