Wednesday, June 4, 2008

Find the latest file in a directory:Bash newbie

I have a huge set of(more than 300) log files(nmn_log.*.txt) in one of my logdir. I had to find out the latest log file name.

$ ls -lrt
total 475244
-rw-r--r-- 1 jks staff 13464 May 20 06:34 nmn_log.23219.txt
-rw-r--r-- 1 jks staff 96851 May 20 06:38 nmn_log.23236.txt
-rw-r--r-- 1 jks staff 96969 Jun 4 11:59 nmn_log.23233.txt
-rw-r--r-- 1 jks staff 91016 Jun 4 12:03 nmn_log.23239.txt
drwxr-xr-x 2 jks staff 4096 Jun 4 12:04 old

The intention is to print the last field of the last line (except the dir old) from the "ls -lrt" output in that directory.

$ ls -lrt | awk '/nmn_log/ { f=$NF };END{ print f }'

So basically the general way of printing the last field of last line(in this case the last field of last line of "ls -lrt" command):

$ ls -lrt | awk '{ f=$NF }; END{ print f }'


Anonymous said...


ls -t1 | head -n1

also do the trick ?

Unknown said...

Paul, you are absolutely right. I did all the awk stuffs just to explain how we can print the last field of last line. Thanks for your comment. Keep reading my blog.


Anthony Lawrence said...

Don't forget that sometimes what you want is the more recently *used*, which "ls -lut" will give you.

Just another handy little thing..

Unknown said...

To get the latest file in a directory tree, use this instead:

find . -not -type d -printf "%T+ %p\n" | sort -n | tail -1

© Jadu Saikia