<!--#exec cgi="list.cgi/elev"--> ???

I need to get the output from one script(WILMA) to get included via SSI to one page.
this script (http://scc.crossnet.se/cgi-bin/kyrktorget/list)
is a linkorganisation script.
The problem is that the categories looks like "list/svk" but
list is NOT a directory but a cgi command(perl).
And "/svk" is just a variable to that script.
My question is. Are there any way to directly include the "list/svk" directly into the webpage or by running it from the prompt?

The problem is that executing "list/svk" from the prompt or SSI means that you should start svk in the directory list.

Can anyone help me?

Linux/ Apache
Who is Participating?
ozoConnect With a Mentor Commented:
<!--#exec cmd="PATH_INFO=/svk; export PATH_INFO; usr/local/etc/httpd/cgi-bin/kyrktorget/list.cgi"-->
What server are you using?
niferAuthor Commented:
Edited text of question
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niferAuthor Commented:
Running a Pentium Linux, with Apache webserver
<!--#exec cmd="list.cgi path=elev"-->
$ENV{PATH_INFO} should be set from the original file,
but you should be able to see @ARGV when called with cmd=
niferAuthor Commented:
How do I fix that?

the source is here:

Maybe something like:

$_ = $1 if( $ARGV[0]=~/\bpath=(\S)*/i );
s|^/*\.*!!;   # strip leading /.
$path_info = $_;

(But I wonder if it might also work to have
<HEAD> <BASE HREF="http://scc.crossnet.se/cgi-bin/kyrktorget/list"> </HEAD>
<!--#exec cgi="list.cgi"-->
and have the path set from the new base?)

On second thought, I don't think <BASE HREF> will work, but you might try something like:
<!--#exec cmd="PATH_INFO=/svk; export PATH_INFO; /usr/local/etc/httpd/cgi-bin/kyrktorget/list.cgi"-->
niferAuthor Commented:
The <!--#exec cmd="PATH_INFO=/svk; export PATH_INFO; /usr/local/etc/httpd/cgi-bin/kyrktorget/list.cgi"-->
Worked Well!!
But the changes of the script was not necessary, (didnt work at all)..

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