I'd like a method to find and print the modified time of a file, for use within a bash script.

I have come up with:

ls -l $filename | cut -d ' ' -f '6-8'

Which outputs:

Jul 26 15:05

Though I'd like to avoid parsing ls, also it'd be useful to have the year in there.

Ideally I'd like to see an output similar to the default output of the date command.

Tue Jul 26 15:20:59 BST 2016

What other useful methods are available?

2 Answers 2


Don't use ls, this is a job for stat:

stat -c '%y' filename

-c lets us to get specific output, here %y will get us the last modified time of the file in human readable format. To get time in seconds since Epoch use %Y:

stat -c '%Y' filename

If you want the file name too, use %n:

stat -c '%y : %n' filename
stat -c '%Y : %n' filename

Set the format specifiers to suit your need. Check man stat.


% stat -c '%y' foobar.txt
2016-07-26 12:15:16.897284828 +0600

% stat -c '%Y' foobar.txt

% stat -c '%y : %n' foobar.txt
2016-07-26 12:15:16.897284828 +0600 : foobar.txt    

% stat -c '%Y : %n' foobar.txt
1469513716 : foobar.txt

If you want the output like Tue Jul 26 15:20:59 BST 2016, use the Epoch time as input to date:

% date -d "@$(stat -c '%Y' a.out)" '+%a %b %d %T %Z %Y'
Tue Jul 26 12:15:21 BDT 2016

% date -d "@$(stat -c '%Y' a.out)" '+%c'               
Tue 26 Jul 2016 12:15:21 PM BDT

% date -d "@$(stat -c '%Y' a.out)"
Tue Jul 26 12:15:21 BDT 2016

Check date's format specifiers to meet your need. See man date too.

  • 6
    Further to this answer, bear in mind that ls is inconsistently implemented across systems, and so should never be used for any sort of automation. Instead, use commands such as stat and find when writing scripts. Aug 2, 2016 at 11:55
  • 2
    excellent. I had not known about this little treasure. So many of them in the *nix/linux environment yet to be discovered.
    – Ken Ingram
    Jan 3, 2019 at 20:37

I tried

ls -l $filename | cut -d ' ' -f '6-8'

but if the date is less than 10 it misses the time. This because of the extra space before the date if less than 10. Try this:

ls -l $filename | awk -F ' ' '{print $6" "$7" "$8}'

The awk command prints the fields separated by all spaces (-F ' '). Hope it works. I know this doesn't answer the original question but just a clarification on the ls command for just date and time. When you Google "ubuntu get date and time of file" it lists this question at the top, which is what I was looking for, since I don't need the year also. For year, date and time, you could try one of the commands below. %m prints month number. %b prints month abbreviation: Drop the %H:%M if you don't need the hour and minute. %-d doesn't print leading zero for the day of the month.

date -r $filename +"%Y %m %-d %H:%M"
date -r $filename +"%y %m %-d %H:%M"
date -r $filename +"%Y %b %-d %H:%M"
date -r $filename +"%y %b %-d %H:%M"

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