'How to exit from a sub menu?
I'm trying to made a option to go out of the program from a sub menu this is my code , but when I use the Exit option in the submenu instead of going out of the program I just go back to the main menu. How can I exit the program from the submenu without using libraries?
def main_menu(input_expected):
while input_expected != "X":
if input_expected=="1":
sub_menu()
else:
print("wrong choice")
print("*** Main Menu ***")
print("----------------------")
input_expected = input(" \
\n [1] Start \
\n [X] Exit \
\n Write your choice: ")
def sub_menu():
new_input = input(" \
\n [1] Continue \
\n [2] Go back \
\n [X] Exit \
\n Write your choice: ")
if new_input =="X":
input_expected = new_input
main_menu(input_expected)
print("*** Main Menu ***")
print("----------------------")
input_expected = input(" \
\n [1] Start \
\n [X] Exit \
\n Write your choice: ")
main_menu(input_expected)
Solution 1:[1]
If you are trying to exit out of the entire script the use sys.exit()
like:
from sys import exit
def main_menu(input_expected):
while input_expected != "X":
if input_expected=="1":
sub_menu()
else:
print("wrong choice")
print("*** Main Menu ***")
print("----------------------")
input_expected = input(" \
\n [1] Start \
\n [X] Exit \
\n Write your choice: ")
def sub_menu():
new_input = input(" \
\n [1] Continue \
\n [2] Go back \
\n [X] Exit \
\n Write your choice: ")
if new_input =="X":
exit()
print("*** Main Menu ***")
print("----------------------")
input_expected = input(" \
\n [1] Start \
\n [X] Exit \
\n Write your choice: ")
main_menu(input_expected)
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Laurinchen |