Re: [xsl] Get Document File name

Subject: Re: [xsl] Get Document File name
From: Abel Braaksma <>
Date: Mon, 23 Oct 2006 08:35:32 +0200
Frank Marent wrote:
hi phil

Is there a way to get the file name of the document you are processing? If I
use Document-uri() it returns the whole file path.

i'm using

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

or '\' if working with backslashes.

A backslash cannot be a literal part of a URI, it is not allowed. It must be escaped for that purpose as %5C. Because a URI may well have additional information after the 'filename', namely the query part and the fragment part, I recommend using the regular expression that is provided as a convenience in the RFC2396 paper. It puts the path in $5::

12            3  4          5       6  7        8 9

You can use the tokenize function above for splitting $5. Not however that that a path separator not necessarily be a slash, it depends on the scheme part ($2) what is and what is not allowed in the path expression.


-- Abel Braaksma

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