防止脚本通过sys.exit（）关闭

Question

I need to call 1 script(test.py) every 5 minutes, so i have used another script timer.py with following code:

import time
while(1==1):
    execfile("test.py")
    time.sleep(300)

This works correctly. But it stopped working after few iterations. After debugging i found that there is a flow in test.py which uses following code:

sys.exit()

So, this is causing both test.py and timer.py to stop. what changes should be done, so as to continue timer.py since i want sys.exit() to only exit test.py and not timer.py

Answer 1

sys.exit() doesn't do more then raising SystemExit (a BaseException subclass), which can be caught like any exception eg:

import time
while True:
    try:
        execfile("test.py")
    except SystemExit:
        print("ignoring SystemExit")
    finally:
        time.sleep(300)

Answer 2

Try this:

import time
import os
while True:
    os.system("python test.py") # if you are not running script from same directory then mention complete path to the file
    time.sleep(300)

Answer 3

You should be able to use:

try:
  # Your call here
except BaseException as ex:
  print("This should be a possible sys.exit()")

Check out the documentation for more information.

Answer 4

Use subprocess

import subprocess  
import time

while(1==1):
    subprocess.call(['python', './test.py'])
    time.sleep(300)

You could even remove the python word if the test.py file has a shebang comment on the first line:

#!/usr/bin/env python

This is not exactly the same, as it will start a new interpreter, but the results will be similar.