integer input validation python

Question

can't seem to get this loop to work, it keeps looping back to the Binary Number Input. I'd like it to loop back to the menu selection. Sorry for the noob question I'm new to python and programming.

import sys
loop = 0
menu_Select = 0
for menu_Select in range(1,100):
    #Display user options to the screen
    print('*** Menu ***')
    print('1. Convert to binary')

    userMenu = input('What would you like to do [1,2,3,4]? ')
    if userMenu != '1' and userMenu != '2' and userMenu != '3' and userMenu != '4':
        print("Please enter either 1, 2, 3, or 4.")

    elif userMenu == '4':
        print('Goodbye.')
        sys.exit(0)

    elif userMenu == '1':
        #Decimal to Binary convertion code
        print('\n')
        while loop < 1:
            while True:
                try:
                    user_Number = (int(input('Please enter number: ')))
                except ValueError:
                    print('wrong')
                else:
                    binary_num = []

                    while (user_Number > 0):
                        if user_Number % 2 != 0:
                            binary_num.append(1)
                        elif user_Number % 2 == 0:
                            binary_num.append(0)
                        user_Number = user_Number // 2
                    binary_num.reverse()
                    binary_display = ''.join(str(k) for k in binary_num)
                    print('Binary number: ',binary_display)
            loop += 1

Something different: You should avoid things like if userMenu != '1' and userMenu != '2' and userMenu != '3' and userMenu != '4': to check for undefined options. Either do if int(userMenu) in range(1,5):, or do the corresponding print command in an else statement after the behavior for valid options is defined. — Bastian35022
– Bastian35022, Commented May 6, 2015 at 8:14
You need a break to exit this loop. Why is that loop in the code anyway? — Matthias
– Matthias, Commented May 6, 2015 at 8:16

SuperBiasedMan · Accepted Answer · 2015-05-06 08:44:42Z

Using input() will actually convert what the user types into an int if it can. So take a look at what happens:

>>> input("= ")
= 12
12

That returns 12, not '12'. For input to give me '12' I need to manually wrap it in quotes.

>>> input("= ")
= '12'
'12'

Instead, use raw_input() to get Python to read anything the user types as a string.

>>> raw_input("= ")
= 12
'12'

Also, as others have mentioned you're using the while loops wrong. If you want to keep asking the user for input until you get a valid number, it's better to wrap a smaller amount of code with the relevant condition.

ie. Have the loop only run while there isn't a valid number, and have it only contain the lines where the input is happening.

        user_Number = None
        while user_Number is None:
            try:
                user_Number = (int(raw_input('Please enter number: ')))
            except ValueError:
                print('wrong')
        binary_num = []

        while (user_Number > 0):
            if user_Number % 2 != 0:
                binary_num.append(1)
            elif user_Number % 2 == 0:
                binary_num.append(0)
                user_Number = user_Number // 2
        binary_num.reverse()
        binary_display = ''.join(str(k) for k in binary_num)
        print('Binary number: ',binary_display)

Bastian35022 · Accepted Answer · 2015-05-06 08:40:06Z

You can introduce a boolean variable done = False, before the while True loop, and change this loop to while not done. Then set done to True after the binary number was printed.

elif userMenu == '1':
    #Decimal to Binary convertion code
    print('\n')
    done = False
    while not done:
        try:
            user_Number = (int(input('Please enter number: ')))
        except ValueError:
            print('wrong')
        else:
            binary_num = []
            while (user_Number > 0):
                if user_Number % 2 != 0:
                    binary_num.append(1)
                elif user_Number % 2 == 0:
                    binary_num.append(0)
                user_Number = user_Number // 2
            binary_num.reverse()
            binary_display = ''.join(str(k) for k in binary_num)
            print('Binary number: ',binary_display)
            done = True

Alex · Accepted Answer · 2015-05-06 08:15:42Z

0

Change:

if userMenu != '1' and userMenu != '2' and userMenu != '3' and userMenu != '4':

To:

if userMenu != 1 and userMenu != 2 and userMenu != 3 and userMenu != 4:

And also update your if statements to see if they are int's rather than strings. This will work on python 2.7, not sure about python 3.

answered May 6, 2015 at 8:15

Alex

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2 Comments

Matthias Over a year ago

The code looks like Python 3 so your answer is wrong. If you read the question carefully you'll see that this is not the problem.

Alex Over a year ago

Gotcha. I put his code into mine which is running 2.7 and ran it, it was looping on the first question and never reaching the if statements. But I see now that he needs to break out of his while True loop too.

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