You could do it using parameters:
Add <xsl:param name="filename"/> to the top of your stylesheet
Access the value of the parameter with: <xsl:value-of select="$filename"/>
Add a 'filename' parameter to the commandline:
saxon -o testdata/out_file.htm ../xml_files/in_file.xml ../styles/test.xsl filename=in_file.xml
-----Original Message-----
From: Encoded Archival Description List [mailto:[log in to unmask]]On Behalf Of
Michele Combs
Sent: 15 February 2007 18:43
To: [log in to unmask]
Subject: Getting name of input file
Anyone know how to access the name of the input file when doing an XSLT
transform? I'm using Saxon, command line as follows:
saxon -o testdata/out_file.htm ../xml_files/in_file.xml
../styles/test.xsl
Within my style sheet (test.xsl) I want to be able to get the name of
the input file, in this case in_file.xml, and output it at a particular
spot in out_file.htm. There's got to be a way but I can't seem to find
it.
Michele C.
-=--=--=--=--=--=--=--=--=--=--=--=--=--=-
Michele R. Combs
[log in to unmask]
Manuscripts Processor
Special Collections Research Center
Syracuse University Library
222 Waverly Avenue
Syracuse, NY 13244
(315) 443-2697
-=--=--=--=--=--=--=--=--=--=--=--=--=--=-
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