I need to remove any file in the directory that is older than 2 years old. It is very important that I keep the newest files and delete the old files.
I have searched and found this.
find /path/to/files* -mtime +365 -exec rm {} \;
Can I just multiply the number?
find /path/to/files* -mtime +1095 -exec rm {} \;
Is there a way to add a switch that will print the file name to the screen as it removes it? To make sure it is doing what I am expecting?
I have also found this:
find /rec -mtime +365 -print0 | xargs -0 rm -f
Is there a major difference between the two? Is one better than the other? What I have read says that xargs is faster. Would I be able to multiply the mtime number out to a 2nd or 3rd year?
And finally would would I be able to place the code as it is into a cron job that can run daily?
Thank you!
Can I just multiply the number?
find /path/to/files -mtime +1095 -exec rm {} \;
Yes. And to "echo" before you remove
find /path/to/files -mtime +1095 -print
Then the version with -exec rm {} \;
to remove the files (when you are ready).
find /path/to/files* -mtime +1095 -exec rm {} \;
That should work fine, you can run a dry a run of this by simply listing the files that are found by the command:
find /path/to/files* -mtime +1095 -exec ls {} \;
To be safe though I would also add in a -type to ensure that other things dont get deleted:
find /path/to/files* -type f -mtime +1095 -exec rm {} \;
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