Why does my if statement return the wrong prompt?-CodePudding

I am sure the answer is right in front of my face but I can't seem to figure out how to fix the return on my first if statement when I input the right conditions.

create_password = input("Enter password here: ")

if len(create_password) > 6 and create_password.isdigit() > 0:
    print("Your account is ready!")
elif len(create_password) < 6 and create_password.isdigit() < 1:
    print("Password must be more than 6 characters and must include a number")
elif len(create_password) > 6 and create_password.isdigit() == 0:
    print("Password must include a number")
else:
    print("Your password sucks")

Let's say I enter elephant100, I am trying to get the prompt to be "Your account is ready!". But to my dismay, it prints "Password must include a number" and I cannot figure out why. My other conditions match the right input but that is the only one that does not work.

CodePudding user response：

.isdigit() method returns True if all the characters are digits, otherwise False. Hence it returns False in this case since your string contains letters like e, l, p etc. So the statement print("Your account is ready!") will never be executed.

CodePudding user response：

Turns out I needed to use isalnum(), isnumeric(), and isalpha() to fix my problem. Thank you Muhammed Jaseem for helping me figure it out! Here is my revised code.

if create_password.isnumeric() == True:
    print("Password must be more than 6 characters and include letters")
elif create_password.isalpha() == True:
    print("Password must be more than 6 characters and include numbers")
elif len(create_password) > 6 and create_password.isalnum() == True:
    print("Your account is ready!")
else:
    print("Your password sucks. Must be more than 6 characters and contain only letters and numbers")