python: input validation return

Question

I am trying to return the number if it is an INT and between numbers, an error occurs when you enter a letter . also you have to input the correct value twice to get an input:

  def get_number():

b = False

while b == False:
    try:
        n = (input('Please enter a 6 digit number'))
    except ValueError:
        continue
    if n >= 100000 and n <= 1000000:
        b = True
        break
return n

if __name__ == '__main__':
    get_number()
    print get_number()

`

Changed input to raw_input , it now work if someone enters a letter. however when i enter the correct input ,it will keep on looping:

def get_number():

b = False

while b == False:
    try:
        n = (raw_input('Please enter a 6 digit number'))
    except ValueError:
        continue
    if n >= 100000 and n <= 1000000:
        b = True
        break
return n

if __name__ == '__main__':
    get_number()
    print get_number()

score 1 · Accepted Answer · answered Oct 21 '15 at 21:34

There are a few problems with your code.

you could just use input to evaluate whatever the unser entered, but that's dangerous; better use raw_input to get a string and try to cast that string to int explicitly
also, if you are using input, then you'd have to catch NameError and SyntaxError (and maybe some more) instead of ValueError
currently, your if condition would allow a 7-digit number (1000000) to be entered; also, you can simplify the condition using comparison chaining
no need for the boolean variable; just break or return from the loop
you call your function twice, and print only the result of the second call

You could try something like this:

def get_number():
    while True:
        try:
            n = int(raw_input('Please enter a 6 digit number '))
            if 100000 <= n < 1000000:
                return n
        except ValueError:
            continue

if __name__ == '__main__':
    print get_number()

python: input validation return

1 Answers1