Python 3.4.2 loops not

Question

t ask the user for input and when they type letters it comes up with an error message (as i want) but even when user enters number it comes up with error. I want it to come out of this loop and on to next for name if correct data is inputed.

Read the [input() docs](https://docs.python.org/3/library/functions.html#input). That function always returns a string, not an integer. — skrrgwasme, Jan 19 '16 at 19:56

furas · Answer 1 · 2016-01-19T19:58:31.273

5

input always returns text - even there are only digits.

Use int(Telephone) to convert to integer or Telephone.isdigit() to check if there are only digits in text.

BTW: both methods are useless if user use spaces or - in phone number ;)

Maybe you will have to remove spaces and - using replace() before you use int() or isdigit()

edited Jan 19 '16 at 19:58

answered Jan 19 '16 at 19:56

furas

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OP, check out [`int()`](https://docs.python.org/2/library/functions.html#int). – Sam Jan 19 '16 at 19:57

AChampion · Answer 2 · 2016-01-19T20:11:50.167

3

You need to convert to an int:

while True:
    try:
        Telephone = int(input("Please enter your telephone number "))
        break
    except ValueError:
        print ("Sorry, system does not recognise input ")

edited Jan 19 '16 at 20:11

answered Jan 19 '16 at 19:58

AChampion

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1

says invalid syntax for 'catch' – Programmer Jan 19 '16 at 20:05
Even using this code and changing catch to except it runs and every time it runs the question just keeps repeating when it should move to next loop – Programmer Jan 19 '16 at 20:11
take a look at the main question i added new bit its get confusing – Programmer Jan 19 '16 at 20:19
@AChampion could you take a look – Programmer Jan 19 '16 at 20:26

Python 3.4.2 loops not

2 Answers2