How can I catch a SIGINT (Ctrl+C) and stop a function called by a FunctionItem?

[thumsl]

Title sums it up. I use a menu to start long running functions and would like to stop them catching a CTRL+C. How can I do this?

[FelipeEmos]

This has nothing to do with this module, this is a python question. I recommend you look up into "Exceptions" and "Error Handling", that's what you are looking for. Every time you press CTRL+C python throws a KeyboardInterrupt exception and your program can catch it and treat it itself (instead of leaving for the normal handling python does).

Use my following code inside your "long running functions" and change the "sys.exit()" for your own thing.
I also recommend you look up "threading", if you are in the "long running functions" game, than probably you shouldn't run them in the same thread as the UI

import sys, time
var = 1

try:
    while(True):
        var += 0.1
        time.sleep(0.1)
except KeyboardInterrupt:
    print("var = ", var)
    sys.exit()

Here are super useful links:

• Python documentation on Exceptions: https://docs.python.org/3/tutorial/errors.html#handling-exceptions
• Python tutorial on Exceptions: https://www.youtube.com/watch?v=NIWwJbo-9_8
• Python tutorial on Threads: https://www.youtube.com/watch?v=IEEhzQoKtQU

Have fun!

[FelipeEmos]

Actually, I've tried to do the handling the way I've described to you and it didn't work....

I don't know why, but I suspect the menu application is catching the exception itself. I stopped investigating because I figured out "Ctrl + D" works fine to kill the current thread and get me back to my Main Menu. Maybe it could work for you if you don't have submenus nested as in my application.

[jamesfowkes]

Hi @FelipeEmos I am looking for an answer to the same issue (capturing Ctrl-C inside a FunctionItem function).
Ctrl-D did not work. Can you help?

[error9900]

I think this is related to this, so adding it here, rather than starting a new Issue:

Simplified code to show the issue I ran into:

This will raise a KeyError and cause an exit, as expected:

from consolemenu import ConsoleMenu
from consolemenu.items import FunctionItem


def raise_exception():
    raise KeyError


menu = ConsoleMenu()
function_item = FunctionItem("Raise KeyError", raise_exception)
menu.append_item(function_item)
menu.show()

This will sometimes flash the Exception text in the terminal, but ultimately returns to the menu:

from consolemenu import ConsoleMenu
from consolemenu.items import FunctionItem

def raise_exception():
    raise KeyError

while True:
    menu = ConsoleMenu()
    function_item = FunctionItem("Raise KeyError", raise_exception)
    menu.append_item(function_item)
    menu.show()

To rule out the while loop somehow causing the issue:
This will raise a KeyError and cause an exit, as expected:

from consolemenu import ConsoleMenu
from consolemenu.items import FunctionItem

def raise_exception():
    raise KeyError

while True:
    raise_exception()

What led me to this was not being able to catch the exceptions raised from menu.show()...:

This makes it seem like menu.show() is dumping the Exception to the terminal and forcing an exit instead of raising it up to the calling function...:

from consolemenu import ConsoleMenu
from consolemenu.items import FunctionItem


def raise_exception():
    raise KeyError


menu = ConsoleMenu()
function_item = FunctionItem("Raise KeyError", raise_exception)
menu.append_item(function_item)
try:
    menu.show()
except KeyError:
    print("Everything is fine!")