
September 12th, 2000, 11:06 AM
|
|
Junior Member
|
|
Join Date: Sep 2000
Location: Milton Keynes, Bucks, England
Posts: 4
Time spent in forums: < 1 sec
Reputation Power: 0
|
|
|
I was hoping I could use a command like
$onefile = 'ls -l $names[$loop]';
where $names[$loop] holds the name of a file
It does not work though. Obviously what I am trying to do is get a listing for a single file.
Anyone tell me why the above line does not work or an alternative method.
Thanks
------------------
|