Error trapping a user input in a function

Question

i have been struggling to use this function to trap the errors from a user input.

def user_choice():
            choice = 4
            while choice > 3
                    choice = int(input("what list is your card in"))
            return choice

any ideas would be appreciated. thanks

What do you mean *trap the error*? Does this help? https://stackoverflow.com/questions/23294658/asking-the-user-for-input-until-they-give-a-valid-response — Chris_Rands, Aug 08 '17 at 09:01
Or are you taking about the missing colon? `while choice > 3:` — Chris_Rands, Aug 08 '17 at 09:03
no its not the colon, it is just how do i get it to ask the question again if the numbers 1,2,3 aren't inputed? — Stanley Shearman, Aug 08 '17 at 09:08
See the first link I provided then, simple sol would be `while choice not in range(1,4):` — Chris_Rands, Aug 08 '17 at 09:09
def user_choice(): choice = 4 while choice not in range(1,4): choice = int(input("what list is your card in")) return choice — Stanley Shearman, Aug 08 '17 at 09:13

score 0 · Answer 1 · answered Aug 08 '17 at 14:20

This can be done simply by implementing an array with the valid arguments you desire then using the in keyword to check if it is there.

def user_choice():
        choice = int(input('what list is your card in?: '))
        valid = [1,2,3]
        if choice in valid:
            return choice
        else: 
            print('oops! must be between 1-3 please try again')
            user_choice()

Error trapping a user input in a function

1 Answers1