title: Comment out and uncomment cron job containing a certain word
body: <p>I have some script running with cron and I need to stop them, so I want to build a shell script that comments out cron jobs containing a certain word.</p>
  
  <p>How can I comment out and uncomment the cron job containing "start_bots" through shell script? I saw some solutions using "sed", but I'm not familiar with it.</p>
  
  <pre><code>*/3 * * * * /home/pi/Development/instabot.py/start_bots.sh
  0 */3 * * * /home/pi/Development/instabot.py/stop_bots.sh
  0 0 */3 * * /home/pi/Development/instabot.py/remove_logs.sh
  </code></pre>