backup_rotate.sh in Bash

Keep the newest three backups, delete the rest, prove it.

#!/usr/bin/env bash
# backup_rotate: keep the newest three backups of a file, drop the rest
set -euo pipefail

work=$(mktemp -d)
trap 'rm -rf "$work"' EXIT
cd "$work"

echo "content" > data.txt

# simulate five dated backups
for stamp in 0501 0502 0503 0504 0505; do
    cp data.txt "data.txt.2026${stamp}.bak"
done

echo "before:"
ls data.txt.*.bak

# newest three survive; ls sorts names, tail slices the rest away
keep=3
ls data.txt.*.bak | head -n -"$keep" | while IFS= read -r old; do
    rm "$old"
    echo "removed $old"
done

echo "after:"
ls data.txt.*.bak
echo "$(ls data.txt.*.bak | wc -l) backups kept"

How it works

1. Dated .bak names sort correctly by name alone.
2. head -n -3 yields everything except the last three.
3. The while-read loop removes and reports each casualty.

Keywords and builtins used here

• cd
• do
• done
• echo
• for
• in
• read
• set
• trap
• while

The run, in numbers

Lines

28

Characters to type

602

Tokens

94

Three-star pace

80 tpm

At the three-star pace of 80 tokens a minute, this run takes about 70 seconds.

Type this snippet

Step 2 of 2 in Encore, step 26 of 26 in Language basics.

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