当用户输入1或2之外的任何其他值时，为什么我的“ else：mainError（）”未执行？ 例如@或a或大于3的任何数字

Question

This is my code.

print("Welcome to the quiz")

print("Would you like to login with an existing account or register for a new account?")

class validation(Exception):

    def __init__(self, error):
        self.error = error

    def printError(self):
        print ("Error: {} ".format(self.error))

def mainError():
    try:
        raise validation('Please enter a valid input')
    except validation as e:
        e.printError()

def login():
    print ("yet to be made")

def register():
    print ("yet to be made")

while True:
    options = ["Login", "Register"]
    print("Please, choose one of the following options")
    num_of_options = len(options)

    for i in range(num_of_options):
        print("press " + str(i + 1) + " to " + options[i])
    uchoice = int(input("? "))
    print("You chose to " + options[uchoice - 1])

    if uchoice == 1:
        login()
        break
    elif uchoice == 2:
        register()
        break
    else:
        mainError()

If I enter 'a', it comes up with this error:

line 35, in <module>
uchoice = int(input("? "))
ValueError: invalid literal for int() with base 10: 'a'

If I enter a number above 2 like '3':

line 36, in <module>
print("You chose to " + options[uchoice - 1])
IndexError: list index out of range

How can I make sure that if a user enters anything other than 1 or 2, it executes my else commands where it calls my mainError() method which contains my exception that the program would display to my user.

Answer 1

The exception is raising because you don't have the options element you're trying to print in the message

 print("You chose to " + options[uchoice - 1])

Here you're trying to get options[a] or options [3], which doesn't exists. Put this print only inside the if/else that has a related option, and another print in the else without one. Something like this:

for i in range(num_of_options):
        print("press " + str(i + 1) + " to " + options[i])
    uchoice = int(input("? "))

    if uchoice == 1:
        print("You chose to " + options[uchoice - 1])
        login()
        break
    elif uchoice == 2:
        print("You chose to " + options[uchoice - 1])
        register()
        break
    else:
        mainError()

Answer 2

uchoice = int(input("? "))

Well here you have to do some error-checking code like:

try:
    uchoice = int(input("? "))
except ValueError:
    <handling for when the user doesn't input an integer [0-9]+>

Then to handle the overflow when a user enters an index which isn't within the list's range:

try:
    options[uchoice - 1]
except IndexError:
    <handling for when the user inputs out-of-range integer>

Of course this adds overhead due to the try: ... except <error>: ... statement so in the most optimal case you would use conditional checking per something like this:

if (uchoice - 1) > len(options):
    <handling for when the user inputs out-of-range integer>