Help
I need this variable $time to show last hour (00..23) So if hour is 00, I want 23. If
hour is 01, I want 00, etc.....
I can run that awk statement on the command line and it works great, but if I try it
in the script it fails with an awk syntax error...
Is there a way to look at last hour with $time1 ?? I would rather keep it all perl.....
#!/usr/bin/perl
use strict;
use POSIX 'strftime';
use warnings;
use Memoize;
my $dir = "/opt/week/";
my $dirfiles = "";
#my $time1 = strftime "%y%m%d%H", localtime;
my $time = system("/bin/date +%H | awk -F '{if ($1 == 00) print 23; else if ($1 <= 10)
print 0$1 -1; else print $1 -1}'");
opendir(DIR, $dir) || die ("Open Dir Failed: $!\n");
my @files = grep(!/^\.\.?$/,readdir(DIR));
my @files = grep /ip.$time/, @files;
closedir(DIR);
foreach $dirfiles (@files)
{
@ARGV = $dir.$dirfiles;
print "editing file $dirfiles\n";
die "Usage: $0 FILES\n" unless @ARGV;
$^I = '.bak'; # turn on in-place editing with backups
while (<>) { # process one line at a time
s{ ^(\d+) }{ convert($1) }ex;
print;
}
print "Done with $dirfiles\n";
}
memoize('convert');
sub convert { strftime('%Y-%m-%d %T', localtime(shift)) }
thanks
rob
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