Can someone please fix it so that the function can detect whether the input has at least one symbol in it?

Question

import os
import time

def strong(password, verifier):
  symbols = "! # $ % & ( ) * + , - . / : ; = ? @ [ ] ^ _ ` { | } ~"
  password = str(password)
  if len(password) > 8:
    if len(password) < 15:
        for i in symbols:
          if password.find(symbols) is True:

I want to see if I can fix this part because I am not sure how I can use this function to find a special character in the password.

            if password.isalnum():
              verifier = 1
              return 'good password', verifier
            elif password.isalpha():
              return 'Your password is only letters'
            elif password.isnumeric():
              return 'Your password is only numbers'
          else:
            return 'no symbols'
    else:
      return 'Too big'
  else:
    return 'Too small'

Does this answer your question? [Regex to Match Symbols: !$%^&\*()\_+|~-=\`{}\[\]:";'<>?,./](https://stackoverflow.com/questions/8359566/regex-to-match-symbols) — QuantumMecha, Dec 17 '21 at 01:18
Does this answer your question? [How to check if a string contains an element from a list in Python](https://stackoverflow.com/questions/6531482/how-to-check-if-a-string-contains-an-element-from-a-list-in-python) — Matthew, Dec 17 '21 at 01:22

score 2 · Answer 1 · answered Dec 17 '21 at 01:16

2

You can use any() along with a generator comprehension to check this:

if any(ch in password for ch in symbols):

I will point out that the symbols list also contains spaces, so you may want to take those out if you don't want a space to qualify as a symbol.

answered Dec 17 '21 at 01:16

John Gordon

29,573
7
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symbols.split() could be used to remove the spaces – Matthew Dec 17 '21 at 01:20

Samwise · Accepted Answer · 2021-12-17T01:45:52.653

You can simplify this (and make debugging easier) by just returning after each failed check instead of having all the deeply nested if/else. Since in the "good password" case you're returning a "verifier" as well as the "good password" message, I'm figuring you probably want to return a false value for the "bad password" cases as well:

def is_strong(password):
    symbols = set("!#$%&()*+,-./:;=?@[]^_`{|}~")
    if len(password) < 9:
        return "Too small", False
    if len(password) > 14:
        return "Too big", False
    if not any(c in symbols for c in password):
        return "No symbols", False
    if password.isalpha():
        return "Your password is only letters", False
    if password.isnumeric():
        return "your password is only numbers", False
    return "Good password", True

Once the function is simplified down, we can see that the isalpha and isnumeric checks won't actually do anything, because we've already established that the password contains non-alphanumeric symbols! Instead I think you want to do:

def is_strong(password):
    symbols = set("!#$%&()*+,-./:;=?@[]^_`{|}~")
    if len(password) < 9:
        return "Too small", False
    if len(password) > 14:
        return "Too big", False
    if not any(c in symbols for c in password):
        return "No symbols", False
    if not any(c.isalpha() for c in password):
        return "No letters", False
    if not any(c.isdigit() for c in password):
        return "No numbers", False
    return "Good password", True

We can test the function like this:

strong, password = False, ""
while not strong:
    password = input("Set password: ")
    msg, strong = is_strong(password)
    print(msg)

Set password: bob
Too small
Set password: bobledeboingerberoo
Too big
Set password: bobbington
No symbols
Set password: bobbington!
No numbers
Set password: bobbington3!
Good password

Thanks! This is nice solution. – Crystallean Dec 17 '21 at 12:40 — Crystallean, Dec 17 '21 at 12:40

score 0 · Answer 3 · answered Dec 17 '21 at 03:14

Make your list of special characters a set and you'll be able to use the isdisjoint function to check that there are no symbols in the password:

if {*"!#$%&()*+,-./:;=?@[]^_`{|}~"}.isdisjoint(password):
    print("Password has no symbols")
else:
    print("Password contains at least one symbol")

Can someone please fix it so that the function can detect whether the input has at least one symbol in it?

3 Answers3