Python Error Handling Retry Code

Question

Im trying to build a definition that does error checking for my definitions. What is the proper way to structure this code for error handling?

I want the script to run a specified definition, if it fails, retry a number of times, then kill it if it times out.

import time

def open_file():
    open('testfile.py')

def run_def(definition):
    for i in range(0, 4):
        try:
            definition
            str_error = None
        except Exception as str_error:
            pass
            if str_error:
                time.sleep(1)
                print(str_error)
            else:
                break
                print('kill script')

run_def(open_file())

I get an error when I try passing the definition into the error check definition. But the script works if I dont put the error checker into a separate definition.

FileNotFoundError: [Errno 2] No such file or directory: 'testfile.py'

I'm not sure what you're trying to do here. Your open_file function returns None. So doing run_def(open_file()) calls run_def with an arg of None. And so definition in the try block won't ever raise an exception. — PM 2Ring
– PM 2Ring, Commented May 16, 2018 at 15:05
Are you perchance running run_def and open_file in different folders? Try using an absolute path to the file. Also, for what you are doing - try looking at decorators realpython.com/primer-on-python-decorators — SteveJ
– SteveJ, Commented May 16, 2018 at 15:15
Im not really opening a file, Im just trying to create a generic error — user9794893
– user9794893, Commented May 16, 2018 at 15:18
Try to remove call on parameter run_def(open_file) then in function do definition(). Normally error will be catched. — Algorys
– Algorys, Commented May 16, 2018 at 15:25
It is unclear what you are trying to do, but I would think run_def should just check the condition of definition and return, you can have a loop that calls it x number of times. — Ywapom
– Ywapom, Commented May 16, 2018 at 15:25

Algorys · Accepted Answer · 2018-05-16 15:34:22Z

1

I'm not sure to understand what you're trying to do, but if you want to catch exception, your call function should be placed in your try / except block.

Like this:

import time

def open_file():
    open('testfile.py')

def run_def(definition):
    for i in range(0, 4):
        try:
            definition()
        except Exception as str_error:
            if str_error:
                time.sleep(1)
                print(str_error)
            else:
                break
                print('kill script')

run_def(open_file)

You don't need to pass after except.

And you don't need to initialize str_error variable before. (Unless you use it before...)

answered May 16, 2018 at 15:34

Algorys

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