How to return to menu by a special key?

Question

Is there any simple method to return to menu by entering a special key? If the user made a mistake while entering the input, user can return to main menu by pressing @ (a special key) key at the end of the input.

x = input("Enter first number: ")

Here, user enter 5, but want to exit by input @. So, x = 5@

I tried some including this

x=input("Enter first number: ")
print(int(a))
if x == "@":
exit()

Also tries this too

for x in ['0', '0$']:
    if '0$' in x:
        break

But, unable to handle numbers and string values together.

Does this answer your question? [How to detect key presses?](https://stackoverflow.com/questions/24072790/how-to-detect-key-presses) — Libra, Jul 21 '22 at 17:05
Not sure I got your question completely right, but you might want to try different approach: `if "@" in str(x): exit()` — Denis Rasulev, Jul 21 '22 at 17:05
Not sure what the final output needs to look like here. Can you show an example? — , Jul 21 '22 at 17:31

score 0 · Answer 1 · answered Jul 21 '22 at 17:07

0

while True:
  x = input()
  if x[-1] == '@':
    continue
  # other stuff

this should work

answered Jul 21 '22 at 17:07

Iresh Sharma

51
4

score 0 · Answer 2 · answered Jul 21 '22 at 17:28

0

Try

user_input = input("Enter your favirote deal:")
last_ch = user_input[-1]
if(last_ch == "@"):
    continue
else:
    //Your Logic

answered Jul 21 '22 at 17:28

Nouman Ahmad

102
7

Ashgabyte · Answer 3 · 2022-07-22T00:11:28.077

0

int(x) will not work if it contains characters other than numbers. use this construction (@ will be ignored by converting a string to int)

x=input("Enter first number: ")
print(int(x if "@" not in x else x[:-1])) # minus last if has @
if "@" in x:
    exit() # or break or return, depends what you want

edited Jul 22 '22 at 00:11

answered Jul 21 '22 at 17:42

Ashgabyte

154
1
5

How to return to menu by a special key?

3 Answers3