[xsl] Get file name without path info

[Nathan Shaw <n8_shaw@xxxxxxxxx>]

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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