Django: Views.py needs to access Class and Function in another python file

Question

in my helper.py I have a class and a function definition:

class Suites():
    NAME_SUITE = ['NATE', 'TOM', 'TED', 'JAN', 'SAM']

def check_suite(name, suite):
    my_suite = Suites[suite] #I want my_suite to = ['NATE', 'TOM', 'TED', 'JAN', 'SAM']
    return name in suite #I want to return True or False

and in views.py I'm trying to find out that user_in_suite is True

 from .helper import check_suite
 from .helper import Suites

 def apps(request):
    user_name = 'NATE'
    suite_to_check = 'NAME_SUITE'
    user_in_suite = check_suite(name=user_name, suite=suite_to_check)

But I'm getting an error for helper.py: 'classobj' object has no attribute '__getitem__

The short answer is name in getattr(Suites(), suite) but that looks like code nightmare to me... what are you actually trying to accomplish here? — Sayse
– Sayse, Commented Sep 19, 2017 at 14:18
I'm just trying to find a way in views.py, to see if a user (which will actually be more like: user_name = str(request.user.name)) is part of a given list (NAME_SUITE = ['NATE', 'TOM',..for example) — NewToJS
– NewToJS, Commented Sep 19, 2017 at 14:23
So why not just do that? why does NAME_SUITE need to be in a class of its own and why do you need to reference it via a string rather than just calling the variable? — Sayse
– Sayse, Commented Sep 19, 2017 at 14:24
I'm going to add more 'suites' to this eventually. I want an extendable way to see if person X is of list Y — NewToJS
– NewToJS, Commented Sep 19, 2017 at 14:26

Sayse · Accepted Answer · 2017-09-19 14:33:44Z

1

def check_suite(name, suite):
    my_suite = Suites.__dict__[suite] # __dict__ gives you access to the dictionary of class.
    return name in my_suite

This should solve your problem.

edited Sep 19, 2017 at 14:33

Sayse

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answered Sep 19, 2017 at 14:22

Amaan Thakur

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