Shell Programming and Scripting

Hi,

This is listing the time when the files are created in the current directory.

I need to check the files created less than 4 hours from the current time

I just found out we could use "240 minutes ago", try this for exactly what you want:

Code:

ls -l . | awk ' BEGIN { "date +\"%Y-%m-%d\" --date=\"240 minutes ago\"" | getline minutes } $6 >= minutes { print $8 } '

---------- Post updated at 02:54 AM ---------- Previous update was at 02:45 AM ----------

ok, now this should work:

Code:

ls -l --time-style=long-iso | awk ' BEGIN { "date +\"%Y-%m-%d\" --date=\"240 minutes ago\"" | getline minutes } $6 >= minutes { print $8 } '

Thanks a lot for you reply.

If am using the above command in the delow directory,
total 59472
-rw-r--r-- 1 an_s users 6920 Jan 30 2009 config.xml
-rw-rw-r-- 1 an_s users 13562 May 24 05:06 build.xml
-rw-r--r-- 1 an_s users 7566860 May 24 05:42 CerD.ear
-rw-r--r-- 1 an_s users 1427 May 24 05:42 log4j.xml
-rw-rw-r-- 1 an_s users 7587840 May 24 05:43 Cer.tar
drwxrwxr-x 3 an_s users 4096 May 24 06:27 may24

I am getting the output as,

05:42
05:43
05:06
05:42
06:27
2009

Here it is just printing the exact time not the files created 4 hours back.

Can you please it....

what did you get when you run "ls -l --time-style=long-iso"?

You could use Perl:

Code:

perl -le'(4/24) > -M and print for @ARGV' *

am gettin this one if i run ls -l

-rw-r--r-- 1 hiran_s users 7566860 May 24 05:42 CerD.ear
-rw-rw-r-- 1 hiran_s users 7587840 May 24 05:43 CerD.tar
-rw-rw-r-- 1 hiran_s users 13562 May 24 05:06 build.xml
-rw-r--r-- 1 hiran_s users 1427 May 24 05:42 log4j.xml
drwxrwxr-x 3 hiran_s users 4096 May 24 06:27 may24
-rw-r--r-- 1 hiran_s users 6920 Jan 30 2009 config.xml

try adding the "--time-style=long-iso" parameter