从另一个脚本访问函数内部的变量

Question

comodin.py

def name():
    x = "car"

comodin_1.py

import comodin

print comodin.x

Error:

Traceback (most recent call last):
  File "./comodin_2.py", line 4, in <module>
    print comodin.x
AttributeError: 'module' object has no attribute 'x'

Is this possible?

Answer 1

In the code you wrote, "x" doesn't exist in "comodin". "x" belongs to the function name() and comodin can't see it.

If you want to access a variable like this, you have to define it at the module scope (not the function scope).

In comodin.py:

x = "car"

def name():
    return x

In comodin_1.py:

import comodin

print comodin.name()
print comodin.x

The last 2 lines will print the same thing. The first will execute the name() function and print it's return value, the second just prints the value of x because it's a module variable.

There's a catch: you have to use the 'global' statement if you want to edit the value "x" from a function (add this at the end of comodin.py):

def modify_x_wrong():
    x = "nope"

def modify_x():
    global x
    x = "apple"

And in comodin_1.py:

print comodin.name()  # prints "car"
comodin.modify_x_wrong()
print comodin.name()  # prints "car", once again
comodin.modify_x()
print comodin.name()  # prints "apple"