Message error by read timeout keeps appearing even though I add 'try except'

import schedule
import subprocess

try:
    subprocess.call("my_code.py", shell=True)
except:
    pass

def trigger():
    try:
        subprocess.call("my_code.py", shell=True)
    except:
        pass
schedule.every(30).seconds.do(trigger)

while 1:
    schedule.run_pending()
    sleep(1)

Inside my_code.py there is a:

response = requests.get(url, headers=headers, timeout=5).json()

So that there was no risk of kill the terminal or showing an error message, I added try except, thinking that this is how would not show errors and would continue looping the next calls.

But this message (along with several other lines with things written) appears:

requests.exceptions.ReadTimeout: 
HTTPSConnectionPool(host='XXXXXXXXXX.com', port=443): 
Read timed out. (read timeout=5)

When it reaches the 5 second limit, how should I proceed so that it doesn't happen since with try except it didn't solve the problem?

Detail, I didn't want to add a try except together to the request in my_code.py, I really would like to know if there is any way to protect my run terminal to these errors but just messing with this code I published.

Solution 1:

Based on the information from: https://docs.python.org/3/library/subprocess.html#older-high-level-api

You should pass subprocess.DEVNULL as value for stderr agument:

import subprocess
subprocess.call(
    "my_code.py",
    shell=True,
    stderr=subprocess.DEVNULL,
    )