Importing another script in python

Question

Let's say I have a this script called script.py:

def num():
    ...
    return num1, num2

def quiz()
   ...
   ...
   num1, num2 = num()   
   name = input('name: ')
   class_num = input('class: ')

if __name__ = '__main__':
    quiz()

In another script I want to use this script and the script.py's defined variables:

import script

script.quiz()

def path(class_num):
    ... 
    # do stuff

This keeps on returning

'NameError: name 'class_num' is not defined'

How can I use all the variable defined in the script.py in my second script?

Answer 1

you defined a function in the script that returns nothing, all the variables that are being defined by user input need to be returned

def quiz():
    stuff
    name = input('...')
    class_num = input('...')
    return [stuff, name, class_num]

the other scripts needs to be changed with

results = script.quiz()

then you parse results they will contain the quiz responses

Answer 2

You can use something like this In your script.py return the class_num variable

def num():
    ...
    return num1, num2

def quiz()
   ...
   ...
   num1, num2 = num()   
   name = input('name: ')
   class_num = input('class: ')
   return class_num


if __name__ = '__main__':
    quiz()

and use it like this

import script

class_num = script.quiz()

def path(class_var = class_num):
    ... 
    # do stuff

in your code you are trying to use a local variable by importing the script.py which is not correct.

hope it may help !

Answer 3

You are trying to make a local variable act as a global one.class_num has a scope within the function quiz, but not outside of it.

In Python, variables that are only referenced inside a function are implicitly global. If a variable is assigned a value anywhere within the function’s body, it’s assumed to be a local unless explicitly declared as global.

The error

'NameError: name 'class_num' is not defined'

Is telling you exactly that, this variable doesn't exist in the scope its called.

To fix the problem simply change

def quiz()
   ...
   ...
   num1, num2 = num()   
   name = input('name: ')
   class_num = input('class: ')
   return class_num

and in your other script catch the return variable

class_num = script.quiz()

def path(class_num):

Answer 4

Python doesn't know what the variable class_num is because it's trapped within your quiz method.

import script

script.quiz()
    ...
    class_num = input('class: ')
    # quiz function exits, making class_num go out of scope

# since class_num doesn't exist, this NameError
def path(class_num):
    ... 

If you want the class name to be usable outside the function, return it

def quiz()
    ...
    ...
    num1, num2 = num()   
    name = input('name: ')
    class_num = input('class: ')
    return class_num  # <=======

And in the other script

class_num = script.quiz()  # <=======

def path(class_num):
    ... 

If you want the other variables from quiz to be available, you have several options:

• Multiple returns, as you did for your num method
• Create something to that holds all your data which counts as a single return. An object would be perfect, other options would be a namedtuple or a dictionary
• Use globals (please don't do this)

Answer 5

I modified the script to show both: a) return value as mentioned by others, b) missing execution of path, with the returned value as parameter value. Here we have script.py

def num():
    num1 = 1
    num2 = 2
    return num1, num2

def quiz():
   num1, num2 = num()
   name = "Andreas"
   class_num = 9
   return class_num

if __name__ == '__main__':
    quiz()

And here something.py

import script

def path(class_num):
    pass
    # do stuff

class_n = script.quiz()
path(class_n)

I hope this can show how it could work. You possibly used the variable class_num somewhere later outside the function path. That gives you a NameError.