Exit script from inside of try-catch block in Python 3

Question

I'm trying to exit a script from inside a try: except: block except it just goes on to the exception case.

None of these...

try:
    exit()
except:
    pass()

try:
    quit()
except:
    pass

import sys
try:
    sys.exit()
except:
    pass

...exit my script, they just go on to the except case.

How would I exit my script from inside one of these blocks?

Tim · Accepted Answer · 2019-07-13 13:23:08Z

3

All of these examples raise the SystemExit exception and you are catching that exception, a blank except clause will catch all exceptions.

This is the reason why you should always specify the exception you intend to catch or at least use except Exception eg

try:
    exit()
except Exception:
    pass

try:
    quit()
except Exception:
    pass

import sys
try:
    sys.exit()
except Exception:
    pass

With that change in place, all of you examples will cause your Python app to exit

answered Jul 13, 2019 at 13:15

Tim

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