How to sort the filenames shown in grep command based on timestamp?

Question

In the following command, how can I get the file names based on its timestamp

$ grep "exit 0" a*
a1.txt:+ exit 0
a2.txt:+ exit 0
a3.txt:+ exit 0
a4.txt:+ exit 0
a5.txt:+ exit 0

$ ls -latr 
-rw-r--r-- 1 user grp  83046 Oct 27 06:46 a5.txt
-rw-r--r-- 1 user grp  68108 Oct 27 07:59 a1.txt
-rw-r--r-- 1 user grp 159792 Oct 27 12:35 a2.txt
-rw-r--r-- 1 user grp 225703 Oct 27 16:41 a3.txt
-rw-r--r-- 1 user grp 246782 Oct 27 20:17 a4.txt

As you see the above output, I am not getting the output based on the timestamp the file was created. Is there any way to achieve it?

Answer 1

If none of the file names contain space, tab, newline (and $IFS has not been modified), ?, *, [ characters or are called -, this should do it:

grep -- "exit 0" $(ls -tr a*)

$() is called command substitution. Your shell will first run ls -tr, then split+glob its output to make up the list of files passed to your grep command.

Answer 2

With the zsh shell, you can affect the order of globs with glob qualifiers.

grep 'exit 0' a*(.Om)

Om is to reverse o⃞rder by m⃞odification time. I also added .⃞ to select only regular files (not directories or pipes or devices or symlinks...).

Answer 3

Thanks Omni. Even the following way, I am able to achieve this. Of course urs is the simple one. thanks a lot

       grep -H "exit 0" /opt/ctmagent/ctm/sysout_archives/141027/d_dms_btch_alrt_ifm_scp_ksh* | awk -F":" '{system("ls -latr "$1"")}'