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I need to delete from a folder all files older than a specific file.
Running bash on CentOS 7.

I have a solution for this, but I think there should be a more elegant way do it:

reference_file=/my/reference/file get_modify_time() { stat $1 | grep -Po "Modify: \K[0-9- :]*" } pit=$(get_modify_time $reference_file) for f in /folder/0000* ; do [[ "$pit" > "$(get_modify_time $f)" ]] && rm $f ; done
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2 Answers 2

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I haven't tried it, but find should be able to handle the whole operation just fine:

$ find dir/ -type f ! -newer reference -delete

... or...

$ find dir/ -type f ! -newer reference ! -name reference -delete

Basically:

  • ! -newer reference matches files which have been modified less recently than reference.
  • -delete deletes them.
  • ! -name reference excludes reference, in case it is also located under dir/ and you want to keep it.

This should delete all files older than reference, and located under dir/.

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    Perfect :). Just added -maxdepth 1 to keep the search not recursive Commented Feb 25, 2015 at 12:33
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    ! -newer means "not newer", so "older or the same age"; the file itself will be matched if it's in the path find looks at, just something to keep in mind. Commented May 14, 2015 at 9:31
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    @ferada It will not if you exclude it, which is why my answer also includes ! -name reference (see the third bullet). Commented May 14, 2015 at 9:34
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compare file modification times with test, using -nt (newer than) and -ot (older than) operators:

if [ "$file1" -ot "$file2" ]; then #do whatever you want; fi
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    Thanks, that's better than my solution but you still need to iterate all files in the folder and compare... The other answer is even more elegant Commented Feb 25, 2015 at 12:36

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