如何从 python 3.8.0 IDLE shell 运行 python 程序

Question

I want to launch the program Tip_and_Tax_Calculator.py from a folder named Python27 in a python 3.8.0 IDLE shell but I want to know which code will work. I have tried the exec function and execFile but they do not work, I don't know if I am using them in the wrong way too.

import Tip_and_Tax_Calculator
exec('C:\Python27\Tip_and_Tax_Calculator.py')

The error I get is:

Traceback (most recent call last): File "C:/Users/Jayan Subramanian/AppData/Local/Programs/Python/Python38-32/Project Launcher.py", line 1, in import Tip_and_Tax_Calculator ModuleNotFoundError: No module named 'Tip_and_Tax_Calculator

Answer 1

you could use the exec function combined with open and read in the following way for python 3.x :

Notice I change from \\ to / :

exec(open('C:/Python27/Tip_and_Tax_Calculator.py').read())

The method you tried and specified is relevant for python 2.x .

Another issue, you do not need to import the Tip_and_Tax_Calculator

EDIT

after the user added the error message the problem that the script didn't run is the fact the module Tip_and_Tax_Calculator is not in the current shell directory, so one need to specify its full/relative path