Whats a good design/approach?

[hshivaraj]

Hi all.

I have this simple code, which does pretty simple things as open a file within the class. But i cant decide what the best design. Is approach1 is better approach2? I need to understand when re-raise an exception and also is it better to catch FileNotFoundError in the class or in the main. What the best approach?

import json import sys class Approach2FileNotFoundException(Exception): pass class Approach1: def __init__(self, config): self.config = self._open(config) def _open(self, config): with open(config, "r") as fh: return json.load(fh) class Approach2: def __init__(self, config): self.config = self._open(config) def _open(self, config): try: with open(config, "r") as fh: return json.load(fh) except FileNotFoundError as err: raise Approach2FileNotFoundException(err) def main(): try: a1 = Approach1("./config.txt") except FileNotFoundError as err: print(err) sys.exit(1) try: a2 = Approach2("./config.txt") except Approach2FileNotFoundException as err: print(err) sys.exit(1) if __name__ == "__main__": main()

Many thanks

[snippsat]

Is approach1 is better approach 2

Both work in this case,approach 2 you can modify the error message.
It's not unclear what happen here in the case without try,except.

def _open(config): with open(config, "r") as fh: return fh.read()

>>> _open('foo.txt') 'hello' >>> _open('foo999.txt') Traceback (most recent call last): File "<stdin>", line 1, in <module> File "file_try.py", line 3, in _open with open(config, "r") as fh: FileNotFoundError: [Errno 2] No such file or directory: 'foo999.txt' [Errno 2] No such file or directory: 'foo999.txt'

Shorter just print the error from last line over.

def _open(config): try: with open(config, "r") as fh: return fh.read() except OSError as err: print(err)

>>> _open('foo.txt') 'hello' >>> _open('foo999.txt') [Errno 2] No such file or directory: 'foo999.txt'