Find string ending with %
- From: gjkeenan@xxxxxxxxxx (Greg John ** CTR ** Keenan)
- Date: Wed, 17 Aug 2005 10:09:45 +1000
Hi,
Need some help with a regex please.
I need to search a file for every instance of a string that ends with %.
Think output from the df command. I need the 1, 2 or 3 digits before the %
passed into the array.
Code:
#!/usr/bin/perl
use warnings;
use strict;
my $outfile2 = "XXX.dat";
open(DATA2, "< $outfile2") || die "Can't open $outfile2: $!\n";
while(<DATA2>) {
my @oput2 = /\b([0-9]+)%/;
print "XXX @oput2 XXX\n";
}
close(DATA2);
It is kind of working but I pick up stuff I don't want e.g:
Input data(XXX.dat):
head
123 123 111% 123
456 22% 456 456 456
789
tail
Output data:
XXX XXX
XXX 111 XXX
XXX 22 XXX
XXX XXX
XXX XXX
How do I tell it not to process every line in the file, just the lines that
contain the relevant string?
Thanks,
Greg Keenan
Lucent Technologies *
Level 1, 68 Waterloo Road
North Ryde NSW 2113
Australia
* +61 2 9491 6898
* gjkeenan@xxxxxxxxxx
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