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List only the recent log files

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Operating Systems AIX List only the recent log files
# 1  
Old 11-11-2009
List only the recent log files

Hi Friends,

I have a list of files in a directory as shown below. It is basically in this format-> yymmdd.hhmmss.filename.out

$ ls -ltr
total 0
-rw-r--r-- 1 sudvishw ghgroup 0 Nov 11 03:58 091104.021020.a.out
-rw-r--r-- 1 sudvishw ghgroup 0 Nov 11 03:59 091104.021110.a.out
-rw-r--r-- 1 sudvishw ghgroup 0 Nov 11 03:59 091104.031020.a.out
-rw-r--r-- 1 sudvishw ghgroup 0 Nov 11 04:00 091104.031110.b.out
-rw-r--r-- 1 sudvishw ghgroup 0 Nov 11 04:00 091104.031505.b.out
-rw-r--r-- 1 sudvishw ghgroup 0 Nov 11 04:00 091104.041555.c.out

I want to list the latest log of each file. ie. the lastest a.out, b.out, c.out, which means I am looking for only the below 3 files out of these 5 files:

All I can think of is have the file names, a.out, b.out, c.out in a input file and have a for loop. The loop would grep for the file name and take the latest one; it is something like,

ls -ltr | grep "a.out" | tail -1
I want to know if we have any other smarter way of doing this without using any loops as the number of files to scan are more.

Last edited by sudvishw; 11-11-2009 at 08:04 AM..
# 2  
Old 11-11-2009
Try this:

ls -lt | 
awk -F "[ .]" 'a[$11]{next}
{a[$11]=1;print $9, $10, $11, $12}' OFS="." | sort -t. -k3

# 3  
Old 11-11-2009
Thank you very much

It worked out fine. I did a small change with the code you gave as shown below:
ls -lt | awk '{print $9}' |awk -F. 'a[$3]{next}{a[$3]=1;print $1, $2, $3, $4}' OFS="." | sort -t. -k3
I am not sure why awk -F"[ .]" did not work. I got just "..." as output when I used it.

Thank you very much for you help. I really appreciate your brilliance.
# 4  
Old 11-11-2009
"ls -lut" might be better (assuming you really want the latest used rather than just latest created)

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