Executing other python scripts gives NameError exception

Question

I have two .py scripts on the same folder, and I have a function like this on one:

def combat_menu():
    exec("combat.py")

Theese scripts are named dungeon.py(the one im executing from) and combat.py.

When I execute this function this error appears:

Exception in Tkinter callback
Traceback (most recent call last):
  File "C:\Users\hp\AppData\Local\Programs\Python\Python38-32\lib\tkinter\__init__.py", line 1883, in __call__
    return self.func(*args)
  File "C:\Users\hp\Documents\piton\Dungeon\dungeon_Data\dungeon.py", line 25, in combat_menu
    exec("combat.py")
  File "<string>", line 1, in <module>
NameError: name 'combat' is not defined

I execute this function from a tkinter button widget like this:

event_button = t.Button(root, text="Open CM", command=combat_menu)

I have tried importing the code, but this gives ImportError.

Please post the relevant code as a [mre] so people (like me ;-) can have a look at it. — Ronald, Jul 23 '20 at 20:32
you probably want to import it. Does this help https://stackoverflow.com/questions/2349991/how-to-import-other-python-files? — Wisa, Jul 23 '20 at 20:37
it's usually advisable to specify the full path to the executing file - sometimes the environment changes the root location in which files are searched for. — Hayden Eastwood, Jul 23 '20 at 22:27

score 1 · Accepted Answer · answered Jul 23 '20 at 20:38

1

The exec() function supports dynamic execution of Python code. It is not executing a shell command. To execute a shell command, in your case calling another python file. Instead you can use for example:

os.system(*cmd*) ->
os.system('combat.py')

answered Jul 23 '20 at 20:38

Ronald

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Executing other python scripts gives NameError exception

1 Answers1