Building a password generator

Question

I have recently built a password generator but wanted to include an aspect where if the user types in a letter instead of a number when defining the length of the password and number of passwords then the output would be to loop back in. If not the password generator would continue if numbers were inputted.

This is my code so far:

import random

char = "abcdefghijklmnopqrstuvwkyzABCDEFGHIJKLMNOPQRSTUVWXYZ123456789!@£$%^&*"

while True:

   
        password_length = input("how long do you want your password? ")
        password_count = input("how many passwords do you want? ")


        if password_length and password_count != type(int):

            print("Please can you enter a number")

        elif password_length and password_count == type(int):

            for x in range(0,int(password_count)):

                password = ""
                    
                for y in range(0,int(password_length)):

                    random_letters = random.choice(char)
                    password += random_letters

                print(password)

Does this answer your question? [Asking the user for input until they give a valid response](https://stackoverflow.com/questions/23294658/asking-the-user-for-input-until-they-give-a-valid-response) — Random Davis, Jan 28 '22 at 16:11
`password_count` is *always* a `str` value; if you want an `int`, you need to call `int` on the value yourself. — chepner, Jan 28 '22 at 16:11
*password_count != type(int)* will always be True because *input()* returns a string object — DarkKnight, Jan 28 '22 at 16:12
It will never *be* the type `int` nor *have* type `int` (which is what it looks like you are trying to check). — chepner, Jan 28 '22 at 16:12

score 1 · Answer 1 · answered Jan 28 '22 at 16:16

1

Try this:

while True:
    try:
        password_length = int(input("how long do you want your password? "))
    except:
        print("Invalid input")
    else:
        break

while True:
    try:
        password_count = int(input("how many passwords do you want? "))
    except:
        print("Invalid input")
    else:
        break

And then you can go on with the rest of your code

answered Jan 28 '22 at 16:16

Riccardo Bucco

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You should never enter two loops, because one may never leave, It's almost always better to specify an explicit exception type. If you use a naked except: clause, you might end up catching exceptions other than the ones you expect to catch - this can hide bugs or make it harder to debug programs when they aren't doing what you expect. – excitedmicrobe Jan 28 '22 at 16:25
@excitedmicrobe Ok so are you saying that I should catch only very specific exceptions? Is there any case in which my code would fail in this scenario? – Riccardo Bucco Jan 28 '22 at 16:28
yes, there's `ValueError` – excitedmicrobe Jan 28 '22 at 16:30
@excitedmicrobe I know there is `ValueError`, but I don't understand why I should use it in this specific case. Which exceptions other than this should I not catch? – Riccardo Bucco Jan 28 '22 at 16:31
I'd suggest a read on [8. Errors and Exceptions](https://docs.python.org/3/tutorial/errors.html) – excitedmicrobe Jan 28 '22 at 16:38
@excitedmicrobe I know that part of the documentation. But thanks for not answering my question. I still not understand in which case my code would give the wrong result. Could you please show me one? Also, if you really care about handling only `ValueError` and not other exceptions, please feel free to edit my answer as you desire :) – Riccardo Bucco Jan 28 '22 at 16:52

excitedmicrobe · Accepted Answer · 2022-01-28T16:42:22.367

0

Python offers a better way to check if a string is a digit or not.

From w3:

The isdigit() method returns True if all the characters are digits, otherwise False.

import random

char = "abcdefghijklmnopqrstuvwkyzABCDEFGHIJKLMNOPQRSTUVWXYZ123456789!@£$%^&*"
while True:
   password_length = input("how long do you want your password? ")
   password_count = input("how many passwords do you want? ")

   if password_count.isdigit() and password_length.isdigit(): 
     password_int = int(password_count)
     password = ""
     for x in range(0,password_int):
         for y in range(0,int(password_length)):
            random_letters = random.choice(char)
            password += random_letters
           
     print(password)
   else: 
      print("Please enter a valid input in numbers")

Output:

how long do you want your password? 8
how many passwords do you want? 1

ABZEG66j

edited Jan 28 '22 at 16:42

answered Jan 28 '22 at 16:18

excitedmicrobe

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Hi Thankyou :), just quickly when I copied your code I had an error which said break is outside the loop, would this mean break goes directly beneath print. – mbv2425 Jan 28 '22 at 16:41
@mbv2425 check the new code. – excitedmicrobe Jan 28 '22 at 16:47
@mbv2425 please check the gray check mark on my answer if you feel it has helped you in any ways. – excitedmicrobe Jan 28 '22 at 16:49
how comes you added this line password_int = int(password_count) after checking if the string was a digit or not? and if so hoe comes you didn't do this for password_length – mbv2425 Jan 28 '22 at 16:52
you already do this in range() – excitedmicrobe Jan 28 '22 at 16:53
Also, for this code if you want more than 1 password it joins the passwords together instead of giving it in separate lines – mbv2425 Jan 28 '22 at 17:16
@mbv2425, I thought that was a bug, I removed that functionality – excitedmicrobe Jan 28 '22 at 17:23
oh no, is the code updated in reflection of receiving more than one password – mbv2425 Jan 28 '22 at 17:26

Building a password generator

2 Answers2