you also could try this
while (<DATA>) {
my $FilePath = (split)[5];
print "\$FilePath = $FilePath \n";
}
__DATA__
30 23 * * * /usr/lbin/sa/sa1 600 6
Michel
-----Message d'origine-----
De: Jas Grewal (DHL UK) [mailto:[EMAIL PROTECTED]
Date: mercredi 13 août 2003 15:29
À: [EMAIL PROTECTED] Org
Objet: 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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