Try
 
$a = "30 23 * * * /usr/lbin/sa/sa1 600 6";
 
$a =~ m|[^\/]+(.*\/[^\/\s]+)\s[^\/]+|i; 
 
$path = $1;
 
print $path ;
 
 
E.
 
LOQUENDO S.p.A. 
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-----Original Message-----
From: Jas Grewal (DHL UK) [mailto:[EMAIL PROTECTED]
Sent: mercoledì 13 agosto 2003 15.29
To: [EMAIL PROTECTED] Org
Subject: regular expresion question.


Hi all,
 
I am trying to write a regular expresion to get a file path from a cron file, I have 
issolated the required lines, but am having problems with extracting just the file 
path and name.
 
The line is as follows :
 
30 23 * * * /usr/lbin/sa/sa1 600 6
 
I inned to extract just the '/usr/bin/sa/sa1' section.
 
Any guidance would be appreciated.
 
Jas
 
 



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