How to show invalid input

Question

My task is:

Write a program that repeatedly prompts a user for integer numbers until the user enters 'done'. Once 'done' is entered, print out the largest and smallest of the numbers. If the user enters anything other than a valid number catch it with a try/except and put out an appropriate message and ignore the number. Enter 7, 2, bob, 10, and 4 and match the output below.

I want to ignore any invalid integer and print 'Invalid Output' message after calculating the maximum and minimum number.But it always prints the invalid message right after user input. How am i supposed to solve this? Thanks in advance.

Code:

largest = None
smallest = None
while True:
    num = input("Enter a number: ")
    if num == "done":
        break
    try:
        n = int(num)
    except:
        print('Invalid input')
    if largest is None:
        largest=n
    elif n>largest:
        largest=n
    elif smallest is None:
        smallest=n
    elif n<smallest:
        smallest=n
print("Maximum", largest)
print('Minimum', smallest)

Answer 1

You could store that invalid output as a boolean variable

largest = None
smallest = None
is_invalid=False
while True:
    num = input("Enter a number: ")
    if num == "done":
        break
    try:
        n = int(num)
    except:
        is_invalid=True
    if largest is None:
        largest=n
    elif n>largest:
        largest=n
    elif smallest is None:
        smallest=n
    elif n<smallest:
        smallest=n
print("Maximum", largest)
print('Minimum', smallest)
if is_invalid:
    print('Invalid Input')