Delete all files except the newest 3 in bash script

Question

Question: How do you delete all files in a directory except the newest 3?

Finding the newest 3 files is simple:

ls -t | head -3
         


        

Answer 1

The following looks a bit complicated, but is very cautious to be correct, even with unusual or intentionally malicious filenames. Unfortunately, it requires GNU tools:

count=0
while IFS= read -r -d ' ' && IFS= read -r -d '' filename; do
  (( ++count > 3 )) && printf '%s\0' "$filename"
done < <(find . -maxdepth 1 -type f -printf '%T@ %P\0' | sort -g -z) \
     | xargs -0 rm -f --

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Explaining how this works:

• Find emits for each file in the current directory.
• sort -g -z does a general (floating-point, as opposed to integer) numeric sort based on the first column (times) with the lines separated by NULs.
• The first read in the while loop strips off the mtime (no longer needed after sort is done).
• The second read in the while loop reads the filename (running until the NUL).
• The loop increments, and then checks, a counter; if the counter's state indicates that we're past the initial skipping, then we print the filename, delimited by a NUL.
• xargs -0 then appends that filename into the argv list it's collecting to invoke rm with.