Deleting an existing job/task in Crontab [duplicate] - linux

Question

This question already has answers here:
Use sed to edit crontab
(2 answers)
Closed 3 years ago.
I'm writing a script and I'm wanting to be able to delete a specific job/task that a user can create using the Crontab command.
I know that to be able to simply delete all jobs/tasks, you just use:
crontab -r;
But if there's multiple jobs/tasks how are you able to list them and then delete selected ones?

Answer 1

Use crontab -e, it should open all the cron tasks in the system editor and then you remove the specific entry and save and exit. Cheers
Edit: Adding remove from script
You can do something like -
crontab -l | grep -v '<SPECIFICS OF YOUR SCRIPT HERE>' | crontab -
from your script. Give it a try and let me know if it worked

Answer 2

Display available jobs with indexing,
read user choice,
delete job by its index
#!/usr/bin/env bash
# Array of cron job entries
typeset -a cron_entries
# Store the contab jobs into an array
mapfile -t cron_entries < <(crontab -l | grep -vE '^(#.*|[[:space:]]*)$')
if (( ${#cron_entries[#]} > 0 )); then
# List all the jobs
echo "Here are the current cron jobs:"
printf 'Index\tJob entry\n'
for ((i=0; i<"${#cron_entries[#]}"; i++)); do
printf '%4d\t%s\n' $i "${cron_entries[i]}"
done
# Prompt user for job index or exit
read -p $'\nPlease choose a job index to delete, or an invalid index to abandon: ' -r answer
# If answer is a positive integer and within array bounds
if [[ "$answer" =~ ^[0-9]+$ ]] && (( answer < ${#cron_entries[#]} )); then
# Show deleted entry
printf '\nDaleting:\t%4d\t%s\n' "$answer" "${cron_entries[answer]}"
# Delete the selected cron entry
unset cron_entries["$answer"]
# Send the edited cron entries back to crontab
printf '%s\n' "${cron_entries[#]}" | crontab -
else
printf '\nAborted with choice %q\nNo job deleted\n' "$answer"
fi
else
printf 'There is no cron job for user: %s\n' "$USER"
fi