Shell Programming and Scripting

If $date is a separate file system, you can consider umount, mkfs, mount.

Is it possible i can run below command on subdirectories in someway and make some loop, but don't understand the logic.

Code:

time perl -e 'for(<*>){((stat)[9]<(unlink))}'

---------- Post updated at 03:08 PM ---------- Previous update was at 03:07 PM ----------

Its new server, already mounted file system 2 days ago.

This isn't faster than rm. rm is not "running slow" here. If you wrote a super-efficient delete program in three assembly-language instructions, it'd still be slow.

The problem is your directory. Every time you delete a file, the kernel has to update the list of files. When the list of files is really big, this can take a long time.

I know this isn't what you want to hear but there's no "delete faster" button.

file sizes are in kilobytes, that is txt files

The issue, as stated before, is the number of files not their sizes.

You said directories have about 2.3G of data and files are kilobytes in size so I'm guessing each directory my have 100,000 or more files.

Unix filesystems are not designed to have such large numbers of files in a single directory and performance is suffering.

how can i run below program in one line. I need to test that.

Code:

use strict;
use warnings;
use File::Find;
my $dir = "/data/2014-01-14/";
find(\&wanted, $dir);
sub wanted {
unlink glob "*.xml";
}

And it was empty, wasn't it?
Redo it, and it will be empty again...