I have a directory containing a high number of files (like logs for every day of the year). I would like to remove all files created before let's say 22/11. How can I achieve that ? Must I use find then exec -rm? I'm using ksh.

  • 1
    The usual caveat is that Unix filesystems don't generally contain a record of when the file was created - only the times that the inode and the content were last modified are available. Oct 6, 2016 at 15:54

3 Answers 3


Using find is still the preferred way of deleting files. See http://mywiki.wooledge.org/UsingFind for more.

One way of doing this is to create a file with the time-stamp in it. e.g

touch -t 201311220000 /tmp/timestamp

Now delete the files GNUfind (assuming in the current directory) that match the time-stamp e.g:

find . -type f ! -newer /tmp/timestamp -delete  

or non GNU find

find . -type f ! -newer /tmp/timestamp -exec rm {} \;
  • 3
    What about this ? Does it work ? find -type f -mtime +5 -exec rm -f {} \; Nov 27, 2013 at 13:56
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    @user1058398 That will delete files older than 5 days from now. Nov 27, 2013 at 14:23
  • Does it also delete folder along with files? @val0x00ff
    – alper
    Jun 15, 2017 at 6:47
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    @Avatar no because the type -f limits the action to the files only. Jun 15, 2017 at 8:20

With GNU or some BSD finds:

find . ! -newermt 2013-11-22 ! -type d -delete

Note that it checks the last modification time of the files. On some BSDs, you can use -newerBt in place of -newermt to check the file's inode birth time if available instead.

Note that it will also delete the files created at 2013-11-22 00:00:00.0000000000 exactly, not that any clock is that precise anyway, but that could cause problems for files whose timestamp has been arbitrary set, such as with touch -d 2013-11-22T00:00:00 some-file (or touch -d 2013-11-22 with some touch implementation). You could always change it to ! -newermt '2013-11-21 23:59:59.999999999999' (GNU) or ! -newermt '2013-11-21 23:59:59' (BSDs, though that would miss the files last modified within the last second of 2013-11-21).

find /path/to/directory/ -mtime +<number of days> -name '<file name>' -exec rm -rf {} \;


find /Netap_fileshare_backup/SQL/DB_backups/xeo/ -mtime +15 -name 'ORA_XEO*' -exec rm -rf {} \;

In this case it will remove all files that start whith "ORA_XEO" with more than 15 days.

  • Though better watch it with the rm -rf, the first command would remove everything contained in any directories that are older than those 15 days. (Also, why the parenthesis around -name?)
    – ilkkachu
    Oct 6, 2016 at 14:01
  • You're right, we dont need to put "(" ")".
    – calafate
    Oct 6, 2016 at 15:00
  • but if we want to add a condition it can be usefull. for example if we want to remove all files that ends with ".jar" or ".cp", and start with "ex". example.jar - it will be removed example.cp - it will be removed example.tar - it wont be removed
    – calafate
    Oct 6, 2016 at 15:03
  • in this case we can use: find /path/to/directory/ -mtime +<number of days> \( -name '*.jar' -o -name '*.cp' \) -name 'ex*' -exec rm -rf {} \;
    – calafate
    Oct 6, 2016 at 15:07
  • That will remove some files from the <number of days> rather than all files older than <number of days> unless you run it exactly at midnight and it finishes within a second. With GNU find you should use the -daystart modifier to coerce the -mtime value to midnight.
    – roaima
    Oct 6, 2016 at 15:58

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