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Re: [xsl] Input Filenames available to the parser?
Subject: Re: [xsl] Input Filenames available to the parser? From: andrew welch <andrew.j.welch@xxxxxxxxx> Date: Wed, 4 Jan 2006 22:30:41 +0000 |
On 1/4/06, Agnisys <agnisys@xxxxxxxxx> wrote: > Hi, > (Using Saxon8B, XSLT 2.0) > Is the name of the input XML file and the XSLT file available to the parser? If not, how can the > following be done? > > For example: > If input is "in.xml" being processed by "xform.xsl", I want the generated output to list the two > files as ... > > : > //Input XML : in.xml > //Transformer : xform.xsl You can use the base-uri() function to return the uri of the tree containing the context node, so to get the input filename use: //Input XML :<xsl:value-of select="tokenize(base-uri(/), '/')[last()]"/> To get the stylesheet filename you could use base-uri(document('')) but do you really need to discover it... couldn't you just hard code it in as you decide the name when you do file -> save :) By the way, there was a change between Saxon 8.6 and 8.6.1 where document-uri() became base-uri() (I think anyway, I had to make the change, maybe Mike will confirm) so it's worth getting the very latest version. cheers andrew
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