Subject: Re: [xsl] function for getting filename From: "andrew welch" <andrew.j.welch@xxxxxxxxx> Date: Fri, 1 Sep 2006 12:27:34 +0100 |
On 01 Sep 2006 12:01:18 +0100, Colin Paul Adams <colin@xxxxxxxxxxxxxxxxxx> wrote:
>>>>> "andrew" == andrew welch <andrew.j.welch@xxxxxxxxx> writes:
andrew> 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.
andrew> In xslt 2.0 I use:
andrew> tokenize(base-uri(.), '/')[last()]
That's not sound - . might not be the document node - if it's an element with an xml:base attribute in scope, then you may get something very spurious indeed.
Regarding xml:base you should get whatever's defined in the xml:base shouldn't you? In which case it's what the XML author intended, and may be what you need to know at that point, depending on your requirements.
If the OP is invoking a transformation from the command line, then he can pass in the file name as a parameter. That is reliable.
Indeed, but in the general case you'll be fine with tokenize(base-uri(.),'/')[last()]
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