So I wrote this code originally for passwordlength(password) to find whether these criterias are met or not...

def passwordlength(password: str):
    upper = digit = special = False
    for char in password:
        if char in "ABCDEFGHIJKLMNOPQRSTUVWXYZ" and "1234567890" and "abcdefghijklmnopqrstuvwxyz":
            upper = True
        if char in "1234567890":
            digit = True
        if char in "$#@%!":
            special = True
        if upper and digit and special:
            return True
    return False

and now I created a new function to find whether the length of the string is within 6 to 12 characters AND whether each element in the string appears lesser than 3 times or not.. and here it is

def passwordOK(password: str):
    if passwordlength(password)== True:
        if len(password) > 6 and len(password) < 12:
            return True
        for char in password:
            if password.count(char) <= 3:
                return True
        else:
            return False
    if passwordlength(password) == False:
        return False
print(passwordOK(password='aaaa1jd4kjfs$jDhfksfd'))

When I tried to run this, it didn't work at all.. Can someone please help me find the error and correct me? Thank You:)

CodePudding user response：

The following would resemble some workable logic and code:

def passwordlength(password: str):
    upper = digit = special = False
    for char in password:
        if char in "ABCDEFGHIJKLMNOPQRSTUVWXYZ" :
            upper = True
        elif char in "1234567890":
            digit = True
        elif char in "$#@%!":
            special = True
    return upper and digit and special

def passwordOK(password: str):
    if passwordlength(password):
        if not (6 < len(password) < 12):
            return False  # return False early
        for char in password:
            if password.count(char) > 3:
                return False # again: return False early 
        return True  # only when all chars have been checked, you know it's ok
    return False

Of course the naming is still weird. If I were to redo it:

from collections import Counter
from string import ascii_uppercase, digits, punctuation

def characterclass_check(password: str):
    for char_class in (ascii_uppercase, digits, punctuation):
        if not any(c in char_class for c in password):
            return False
    return True

def passwordOK(password: str):
    if not characterclass_check(password):
        return False
    if not (6 < len(password) < 12):
        return False
    return max(Counter(password).values()) <= 3

CodePudding user response：

There are many problems with this code.

First

if char in "ABCDEFGHIJKLMNOPQRSTUVWXYZ" and "1234567890" and "abcdefghijklmnopqrstuvwxyz"

will be evaluated as

if (char in "ABCDEFGHIJKLMNOPQRSTUVWXYZ") and ("1234567890") and ("abcdefghijklmnopqrstuvwxyz")

thus only checking if the character is uppercase, as the other strings will evaluate to True simply by having content. The variable name you're setting is upper, so I'm not sure why those other strings are there to start with - they're not what you're trying to check against.

Second, you unnecessarily call passwordlength(password) twice when the second one could (and should) be an else.

Third, you return immediately on checking the password length in passwordOK, so if it's between 6 and 12 characters you never check for duplicate characters.

Fourth, when you check for duplicate characters you return as soon as you find one that appears less than 4 times, meaning that "aaaaaabbbbbbcdddddd" would pass. I assume this was not your intention.

Fifth, your else on the for loop is unnecessary as in your case it will always run if you don't return from within the loop. It's only useful if you have a break.

Sixth, you have a function named passwordlength that doesn't actually check the length of the password.

I may have missed other issues.