CODEWITHSUNDEEP
pythonpython-3.xasciiord
Ammar Khazal
Ammar Khazal

Reputation: 1

How do I get the ASCII value of a string that is called upon by a list in python 3?

So here is my code (mind the tiny syntax mistakes:

UserID = input("Please enter your UserID ")

if len(UserID) !=6:
    print("Wrong Format")

elif UserID[:1] == (UserID[:1]).lower():
    print("Wrong Format")

elif UserID[1:3] == (UserID[1:3]).upper():
    print("Wrong Format")

elif UserID[3:] > ord(UserID[3:]):
    print("Wrong Format")

else
    print("Correct Format")

Basically, the purpose of this program is to have a UserID of 6 characters with one upper case letter, two lower case letters and 3 digits in the format

Abc123

I run into an issue here

elif UserID[3:] > ord(UserID[3:]):
    print("Wrong Format")

where the ord() function cannot evaluate the ASCII equivalent of a list. I know it is supposed to be for characters so I am stuck as to what to do. This section of the code is to ensure that any number from the 3rd element onwards is a number so it is less than the ascii equivalent of 9.

Upvotes: 0

Views: 81

Answers (3)

Joachim Isaksson
Joachim Isaksson

Reputation: 180887

Just for reference, here's how you'd validate the whole username with a regex;

import re

if re.match('^[A-Z][a-z]{2}[0-9]{3}$', UserID):
    print("Correct Format")
else:
    print("Wrong Format")

In your existing code, to check they're all numeric you don't need ord, you can just compare all characters that they're between 0 and 9 inclusive;

if not all(c >= '0' and c <= '9' for c in UserID[3:]):
    print("Wrong format")

Upvotes: 1

Chad S.
Chad S.

Reputation: 6633

Could just be simplified to: 

UserID = input("Please enter your UserID ")

if ( (len(UserID) !=6) or UserID[0].islower() 
     or UserID[1:3].isupper() or not UserID[3:].isdigit() ) :
    print("Wrong Format")

else
    print("Correct Format")

Upvotes: 0

zondo
zondo

Reputation: 20336

To see if a string is composed only of integers, you can use str.isdigit():

elif not UserID[3:].isdigit():
    print("Wrong Format")

Apparently, (from the comments), there are some things for which str.isdigit() returns True even if it isn't an integer. To fix that, do this:

elif not all(c in "0123456789" for c in UserID[3:]):
    print("Wrong Format")

Upvotes: 1

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