Programming

Hi,
plz see the below code , i have catch the file "Orders.20110714.out "file
as a Orders*.out. but it giving me an error .it does not open the file.

if the same thing i have done by code

code-> ls Orders*.out then it gives me the output
Orders.20110714.out

i am trying apply the same concept ne below code but it is not working

can anyone help me regarding this???

==============================================

Code:

#!/usr/bin/perl -w
$ARGV[0]||='/home/user01/Orders*.out';
$linenum=0;
open OR,$ARGV[0];
local $/="\r";
while (<OR>) {
$linenum++;
}
#close OR;
print $linenum;
close OR;

===============================================

Regards,
Priyanka

---------- Post updated at 02:14 PM ---------- Previous update was at 11:03 AM ----------

Hi all, need ur help

Use file globbing (<*.out>) while opening the file ( read more about glob in perl )

Code:

open OR,</home/sathiya/perl/Orders*.out>;

Not an excellent method to do so, but it works.