[xslt] xsl:variable or param for argv ?



Hi,
Is there a way to access input file name for the xsl ?
I mean something like  :
 % xsltproc t.xsl in.xml

=> value-of select="$inputfile" => "in.xml"

thanks for your help,



	

	
		
__________________________________________________________________ 
Découvrez le nouveau Yahoo! Mail : 250 Mo d'espace de stockage pour vos mails ! 
Créez votre Yahoo! Mail sur http://fr.mail.yahoo.com/


[Date Prev][Date Next]   [Thread Prev][Thread Next]   [Thread Index] [Date Index] [Author Index]