Write Your Own Method of Script using the below scenario and comment. Recently i came across this situation where i could not able to delete more than x days data using the find command due to -mtime is not matching.

Delete More than x Days Data

Scenario: Every day from Monday to Friday one directory will be created under /fullbackup/dailybackup/YYYY-MM-DD and it will move same dirctory to its parent directory everyday midnight /fullbackup/archive/, However Saturday, Sunday and Monday directories will move to /fullbackup/archive path every Monday evening.

Directory Names Example:

$ ls -ltrdrwxrwxrwx 2 ravi backupgroup 4096 Dec 29 12:25 2018-12-28
drwxrwxrwx 2 ravi backupgroup 4096 Dec 29 12:25 2018-12-29

Question: I would like to delete directories older than two days from /fullbackup/archive path. How do you do it using any scripting methods?

Problem Statement: I was trying to use find . -type d -mtime +4 -exec rm -rf {} \;. This command does not work because the modified date for all moved directories on Monday is the same. I can’t differentiate the last three directories modified date which is -mtime.

How Do you solve it.?? Write a Shell Script to accomplish this task. Should run through crontab and clear directories older than two days.

Shell Script to Delete More than x Days Data

#!/usr/bin/env bash

set -euo pipefail

# -----------------------------
# Configuration
# -----------------------------
ARCHIVE_DIR="/fullbackup/archive"
LOG_FILE="/scripts/joblog.log"
RETENTION_DAYS=2 # Change X days here
DRY_RUN=false # Set true to test without deleting

# Directory name date format expected:
# Example: 2026-05-01
DATE_FORMAT_REGEX='^[0-9]{4}-[0-9]{2}-[0-9]{2}$'

# -----------------------------
# Logging function
# -----------------------------
log_msg() {
echo "$(date '+%Y-%m-%d %H:%M:%S') - $*" >> "$LOG_FILE"
}

# -----------------------------
# Validation
# -----------------------------
if [[ ! -d "$ARCHIVE_DIR" ]]; then
log_msg "ERROR: Archive directory does not exist: $ARCHIVE_DIR"
exit 1
fi

CURRENT_EPOCH=$(date +%s)
RETENTION_SECONDS=$(( RETENTION_DAYS * 24 * 60 * 60 ))

log_msg "Backup cleanup started. Retention: ${RETENTION_DAYS} days. Dry run: ${DRY_RUN}"

# -----------------------------
# Process directories
# -----------------------------
find "$ARCHIVE_DIR" -mindepth 1 -maxdepth 1 -type d -print0 |
while IFS= read -r -d '' DIR_PATH; do

DIR_NAME=$(basename "$DIR_PATH")

# Validate directory name format
if [[ ! "$DIR_NAME" =~ $DATE_FORMAT_REGEX ]]; then
log_msg "Skipping invalid directory name: $DIR_PATH"
continue
fi

# Convert directory name to epoch
if ! DIR_EPOCH=$(date -d "$DIR_NAME 00:00:00" +%s 2>/dev/null); then
log_msg "Skipping invalid date directory: $DIR_PATH"
continue
fi

AGE_SECONDS=$(( CURRENT_EPOCH - DIR_EPOCH ))
AGE_DAYS=$(( AGE_SECONDS / 86400 ))

if (( AGE_SECONDS > RETENTION_SECONDS )); then
if [[ "$DRY_RUN" == true ]]; then
log_msg "DRY RUN: Would delete $DIR_PATH, age: ${AGE_DAYS} days"
else
log_msg "Deleting $DIR_PATH, age: ${AGE_DAYS} days"
rm -rf -- "$DIR_PATH"
fi
else
log_msg "Keeping $DIR_PATH, age: ${AGE_DAYS} days"
fi

done

log_msg "Backup cleanup completed."

Explanation

• ls -ltr line will list all the directories on the full backup path and print 9th column using awk and store output to file
• For loop is using output file and prints one directory name YYYY-MM-DD 
• STARTTIME will convert directory name to EPOC time by adding 00:00:00 Hours, Minutes, and Seconds
• ENDTIME prints current date & time in EPOC time format
• echo statement will minus End time – start time and print value into minutes
• COUNT variable is assigned with minutes value
• if statement will validate value is more than 2880 minutes means more than two days of time
• if the value is more than 2880 then rm will delete the directory

Solve the scenario using your own method and post the solution in a comment. Thanks.

Note: Above script works only when the directory name is Year-Month-Date otherwise it will give an error message.

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Ravi Kumar Ankam

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