使用crontab -e，它将在系统编辑器中打开所有cron任务，然后删除特定的条目并保存并退出。干杯

编辑：添加从脚本中删除

您可以执行类似-

crontab -l | grep -v '<SPECIFICS OF YOUR SCRIPT HERE>' | crontab -

来自您的脚本。试试看，让我知道它是否有效

显示带有索引的可用作业， 阅读用户的选择， 按其索引删除作业

#!/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