CODEWITHSUNDEEP
pythonpython-3.xerror-handling
user9794893
user9794893

Reputation: 133

Python Error Handling Retry Code

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'

Upvotes: 0

Views: 480

Answers (1)

Algorys
Algorys

Reputation: 1790

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...)

Upvotes: 1

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