[SOLVED] Removing files older than X days using a date format in the filename

Issue

I have a directory that includes log files, that includes the following date format (YYYY-MM-DD) in the filename.

app-2020-12-17.log.2
app-2020-12-18.log.1
app-2020-12-18.log.2
app-2020-12-18.log.31
app-2020-12-18.log.32
app-2020-12-18.log.33
app-2020-12-18.log.3.gz
app-2020-12-20.log.4.gz
app-2020-12-20.log.5

My goal is to remove all files older than X date (using the date in the filename as comparison). For example, I want to remove all files older than 2020-12-20, so after the operation I should be left with the following files in my folder:

app-2020-12-20.log.4.gz
app-2020-12-20.log.5

How can I achieve this behavior? I have tried using somethine like this find app*log* -maxdepth 1 -mtime +1, but this only finds files by modification date, and not the way I need.

Solution

With bash and a regex:

for i in app-*.log*; do
  [[ "$i" =~ -([0-9]{4}-[0-9]{2}-[0-9]{2}) ]] \
    && [[ "${BASH_REMATCH[1]}" < "2020-12-20" ]] \
    && echo rm -v "$i"
done

${BASH_REMATCH[1]} contains 2020-12-17, e.g.

As one line:

for i in app-*.log*; do [[ "$i" =~ -([0-9]{4}-[0-9]{2}-[0-9]{2}) ]] && [[ "${BASH_REMATCH[1]}" < "2020-12-20" ]] && echo rm -v "$i"; done

Output:

rm -v app-2020-12-17.log.2
rm -v app-2020-12-18.log.1
rm -v app-2020-12-18.log.2
rm -v app-2020-12-18.log.31
rm -v app-2020-12-18.log.32
rm -v app-2020-12-18.log.33
rm -v app-2020-12-18.log.3.gz

Answered By - Cyrus

Answer Checked By - Cary Denson (WPSolving Admin)