Subject: [xsl] Get file name without path info From: Nathan Shaw <n8_shaw@xxxxxxxxx> Date: Fri, 8 Mar 2002 10:47:51 -0800 (PST) |
I am using saxon:system-id() to get the name of the xml file being parsed. However, as you probably know, Saxon returns the entire path along with the file name. I want to just get the file name, strip off the '.xml' and replace it with '_p.html'. So, Saxon gives me this: file:/C:/Documents and Settings/nshaw.HQIRMS/My Documents/spaceresearch/newxml/general_info/what.xml and I want to end up with this: what_p.html I am feeling rather handicapped by XSL when it comes to doing this. I am used to having functions like split(), gettoken(), replace(), etc... available to me. Can someone give me some guidance as to how to do this in XSL? Thanks, --Nate __________________________________________________ Do You Yahoo!? Try FREE Yahoo! Mail - the world's greatest free email! http://mail.yahoo.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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