In if-else statement throwing this error "ValueError: invalid literal for int() with base 10: ''

Question

I was trying to build a password generator in Python with this code:

import random
import string

print("Welcome to password generator !")

def password_generator():


    lowercase_data = string.ascii_lowercase

    uppercase_data = string.ascii_uppercase
    numbers = string.digits
    symbols = string.punctuation

    combined = lowercase_data+uppercase_data+numbers+symbols

    lenght= int(input("Enter your password lenght:"))

    if  lenght == None:
        print("please enter your value")
        password_generator()

    else:


        generator = random.sample(combined,lenght)

        password = "".join(generator)

        print(password)
        password_generator()


password_generator()

When my input is blank it shows this error:

ValueError: invalid literal for int() with base 10: ''

Now what should I do with this code??

If my input is blank then I want to continue with the if-statement.

Answer 1

You need to catch an exception here. You try to parse a string to int by using int(...) on a string that can literally contain anything.

In your case, it is empty, so it cannot parse anything to an integer. Your if length == None afterwards is rather useless, because int() will not return None but instead throw an error (see the doc).

You need to catch this error one way or the other, either by surrounding the statement with a try except block or by sanitizing the input (i.e. taking a = input("Enter your pw length:") and then checking if it only contains digits before you call int(a)

Answer 2

try:
    lenght = int(input("Enter your password lenght:"))
except ValueError:
    lenght = None

if  lenght is None:
    ...

Trying to convert a non number value to an int generates a ValueError - as you've seen. If that happens, set lenght to None, and carry on.