I am writing a script that:

1. Asks the user to enter a number.
2. Detects errors of type 'ValueError'. The user has only 5 attempts to enter correct input.
3. If the user types in zero, the script stops execution with the message 'All done'.
4. If the user enters a positive integer, the script tells if the number is even or odd. It continues asking the user to type in another number; 10 times in total.

Here is my attempt.

attempt = 0
max_attempts = 5
i = 0
i_max = 10

while attempt < max_attempts:
    attempt  = 1
    try:
        while i < i_max:
            i  = 1
            number = int(input('Please type in a positive integer; 0 to quite: '))
            if number == 0:
                print('All done!')
                break
            if number <0:
                print('Please type in a positive number.')
            if number%2 == 0:
                print("The number",number, "is even.")
            else:
                print("The number",number, "is odd.")
            print('Only', i_max-i, 'numbers left.')
    except ValueError:
        print('Wrong input, ', max_attempts - attempt, 'attempts left')
if attempt == max_attempts: # You tried too many times
    print('Sorry, too many attempts!')

If the user enters 0 the script does not cease as desired. If the user enters correct input the final string 'Sorry, too many attempts!' is printed which is not desired in this case. How I can solve these issues? Any help will be greatly appreciated. Thank you very much in advance. I apologize if I am missing something rather elementary.

CodePudding user response：

If the user enters 0, then you break the current while loop. However, this while loop is in another while loop. You can add a variable "is_finish". If the user enters 0, set this new variable to True. In the first while loop condition, add "is_finish == False".

Hope this help you !

CodePudding user response：

You can use a for loop to control the total number of repeats the user is allowed, then a separate failure variable to count the number of times they enter something invalid:

max_success = 10
max_failures = 5

failure = 0
for attempt in range(max_success):
    user_input = input(f' {attempt  1}/{max_success} - Ener a positive integer or 0 to quit: ')
    if user_input == '0':
        print(' Quit')
        break
    try:
        user_input = int(user_input)
        if user_input % 2 == 0:
            print(' Even number')
        else:
            print(' Odd number')
    except ValueError:
        failure  = 1
        print(f' Not a number (failure {failure} of {max_failures})')
        if failure >= max_failures:
            break

Here's an example output if they submit 5 invalid entries:

 1/10 - Ener a positive integer or 0 to quit: 5
 Odd number
 2/10 - Ener a positive integer or 0 to quit: a
 Not a number (failure 1 of 5)
 3/10 - Ener a positive integer or 0 to quit: b
 Not a number (failure 2 of 5)
 4/10 - Ener a positive integer or 0 to quit: c
 Not a number (failure 3 of 5)
 5/10 - Ener a positive integer or 0 to quit: d
 Not a number (failure 4 of 5)
 6/10 - Ener a positive integer or 0 to quit: e
 Not a number (failure 5 of 5)

CodePudding user response：

add attempt = 6 inside the if condition (if number == 0)

attempt = 0
max_attempts = 5
i = 0
i_max = 10

while attempt < max_attempts:
    attempt  = 1
    try:
        while i < i_max:
            i  = 1
            number = int(input('Please type in a positive integer; 0 to quite: '))
            if number == 0:
                print('All done!')
                attempt = 6
                break
            if number < 0:
                print('Please type in a positive number.')
            if number % 2 == 0:
                print("The number", number, "is even.")
            else:
                print("The number", number, "is odd.")
            print('Only', i_max - i, 'numbers left.')
    except ValueError:
        print('Wrong input, ', max_attempts - attempt, 'attempts left')
if attempt == max_attempts:  # You tried too many times
    print('Sorry, too many attempts!')