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Get file name without path info
- From: Nathan Shaw <n8_shaw at yahoo dot com>
- To: XSL-List at lists dot mulberrytech dot com
- Date: Fri, 8 Mar 2002 10:47:51 -0800 (PST)
- Subject: [xsl] Get file name without path info
- Reply-to: xsl-list at lists dot mulberrytech dot com
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
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