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I was reading a similar question Returning error string from a function in python. While I experimenting to create something similar in an Object Oriented programming so I could learn a few more things I got lost.

I am using Python 2.7 and I am a beginner on Object Oriented programming.

I can not figure out how to make it work.

Sample code checkArgumentInput.py:

#!/usr/bin/python

__author__ = 'author'


class Error(Exception):
    """Base class for exceptions in this module."""
    pass


class ArgumentValidationError(Error):
    pass

    def __init__(self, arguments):
        self.arguments = arguments

    def print_method(self, input_arguments):
        if len(input_arguments) != 3:
            raise ArgumentValidationError("Error on argument input!")
        else:
            self.arguments = input_arguments
            return self.arguments

And on the main.py script:

#!/usr/bin/python
import checkArgumentInput

__author__ = 'author'


argsValidation = checkArgumentInput.ArgumentValidationError(sys.argv)

if __name__ == '__main__':

    try:
        result = argsValidation.validate_argument_input(sys.argv)
        print result
    except checkArgumentInput.ArgumentValidationError as exception:
        # handle exception here and get error message
        print exception.message

When I am executing the main.py script it produces two blank lines. Even if I do not provide any arguments as input or even if I do provide argument(s) input.

So my question is how to make it work?

I know that there is a module that can do that work for me, by checking argument input argparse but I want to implement something that I could use in other cases also (try, except).

Thank you in advance for the time and effort reading and replying to my question.

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2 Answers 2

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OK. So, usually the function sys.argv[] is called with brackets in the end of it, and with a number between the brackets, like: sys.argv[1]. This function will read your command line input. Exp.: sys.argv[0] is the name of the file.

main.py 42

In this case main.py is sys.argv[0] and 42 is sys.argv[1].

You need to identifi the string you're gonna take from the command line. I think that this is the problem.

For more info: https://docs.python.org/2/library/sys.html

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3 Comments

Hello TheShadlest, I will get back to you on this one later on. Unfortunately I do not have access on the script right now. Again thank you for your time and effort reading and replying to my question.
Hello again TheShadlest, apologies for so late reply but I got busy with something else. Unfortunately your answer is not correct. Well you are right about the sys.argv[0] and sys.argv[1] but my code had many errors. I spend some time and I have found the solution. I will be posting the answer to my question actually now, so someone in future might benefit from that.
I actually will benefit from it, thanks. I say, will help me to help others in the future. Thanks for replying.
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I made some research and I found this useful question/ answer that helped me out to understand my error: Manually raising (throwing) an exception in Python

I am posting the correct functional code under, just in case that someone will benefit in future.

Sample code checkArgumentInput.py:

#!/usr/bin/python

__author__ = 'author'


class ArgumentLookupError(LookupError):
    pass

    def __init__(self, *args): # *args because I do not know the number of args (input from terminal)
        self.output = None
        self.argument_list = args

    def validate_argument_input(self, argument_input_list):
        if len(argument_input_list) != 3:
            raise ValueError('Error on argument input!')
        else:
            self.output = "Success"
            return self.output

The second part main.py:

#!/usr/bin/python
import sys
import checkArgumentInput

__author__ = 'author'

argsValidation = checkArgumentInput.ArgumentLookupError(sys.argv)

if __name__ == '__main__':

    try:
        result = argsValidation.validate_argument_input(sys.argv)
        print result
    except ValueError as exception:
        # handle exception here and get error message
        print exception.message

The following code prints: Error on argument input! as expected, because I violating the condition.

Any way thank you all for your time and effort, hope this answer will help someone else in future.

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