CODEWITHSUNDEEP
pythoncronos.system
mjoy
mjoy

Reputation: 680

Os.system doesn't push message in cron alert to cronitor?

I am running a python script using a cron job. When the script runs, it either alerts that it completed or failed. It is supposed to send a message to cronitor along with the completion ping. An example of the link is below:

 https://cronitor.link/p/id/key?state=complete&msg='Success'

When I just put the link into the search bar and hit Enter, the message shows up in cronitor. However, when I try to get the message to print when running the link through the script it doesn't work. Cronitor gets the ping that it completed successfully but no message shows up:

 cron_alert = "/usr/local/bin/curl --silent https://cronitor.link/p/id/key?state=complete&msg='Success'"
 os.system(cron_alert)

I tried removing '--silent' but that didn't make a difference. Any ideas on how to fix this? Thanks

Upvotes: 0

Views: 228

Answers (1)

chepner
chepner

Reputation: 531165

Your command contains an unescaped &, which the shell used by os.system parses as a command terminator that runs curl the the background. You need to escape it, e.g.

 cron_alert = \
   "/usr/local/bin/curl --silent \"https://cronitor.link/p/id/key?state=complete&msg='Success'\""

but even better would be to stop using os.system and use subprocess.run instead.

import subprocess


subprocess.run(["/usr/local/bin/curl",
                "--silent",
                "https://cronitor.link/p/id/key?state=complete&msg='Success'"])

which bypasses the shell altogether.

Best, use a library like requests to send the request from Python, rather than forking a new process.

import requests

requests.get("https://cronitor.link/p/id/key",
              params={'state': 'complete', 'msg': "'Success'"})

(In all cases, does the API really require single quotes around Success?)

Upvotes: 1

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