check if string contains an integer python

Question

I need to check if a character entered in a password contains an integer or not.

password = input("Enter a password:" )
for num in password:
    if num.isdigit():
        break
    else:
        print("Your password must contain a number.")

The code above doesn't work because I'm assuming due to python 3 taking every user input as a string, it checks the string and never knows the difference between the string and the integer in the string. How would I get around this?

Does this answer your question? [Check if a string contains a number](https://stackoverflow.com/questions/19859282/check-if-a-string-contains-a-number) — Alexander Santos, Dec 14 '21 at 19:54
You give up checking and proclaim it must contain a number right on the first character. Remove the `else` and print/handle failure only after the entire loop is done, meaning you checked each character and none were digits (the `break` should probably set some sort of variable first). Simpler: `any(x.isdigit() for x in password)` — ggorlen, Dec 14 '21 at 19:56
Or did you intend the `else` belong to the `for` instead of the `if`? That would actually be correct. — mkrieger1, Dec 14 '21 at 19:58
Thanks everyone, I've realised it's a silly indentation error and just completely overlooked it. Yes I did intend for that @mkrieger1 — korflol, Dec 14 '21 at 20:37

score 3 · Accepted Answer · answered Dec 14 '21 at 20:04

Your code works fine if you unindent the else to make it part of the for:

for num in password:
    if num.isdigit():
        break
else:
    print("Your password must contain a number.")

If it's part of the if, the else happens for every character that's not a digit; if it's part of the for, it happens at the end of the loop if the loop was never broken, which is the behavior you want.

An easier way of writing the same check is with the any function ("if there aren't any digits..."):

if not any(num.isdigit() for num in password):
    print("Your password must contain a number.")

or equivalently with all ("if all the characters aren't digits..."):

if all(not num.isdigit() for num in password):
    print("Your password must contain a number.")

Thank you, silly mistake. I may adopt the any function method. — korflol, Dec 14 '21 at 20:39

Mohammad Khoshbin · Answer 2 · 2021-12-14T20:09:29.917

Your code will print Your password must contain a number. for every non-digit character. You need to know if there is a digit in the password or not, then make decision about its validity. So you can use a boolean variable, then if any digit character found in the password, the variable would be True and after for loop you will use it for checking the validity of password.

You can do like this:

password = input("Enter a password:" )
good_pass = False

for num in password:
    if num.isdigit():
        good_pass = True
        break

if good_pass == True:
    print('Good password!')
else:
    print("Your password must contain a number.")

thanks, solved it. Didn't realise my error. – korflol Dec 14 '21 at 20:40 — korflol, Dec 14 '21 at 20:40

check if string contains an integer python

2 Answers2