CODEWITHSUNDEEP
pythonpython-3.x
user4267777
user4267777

Reputation:

Python - invalid literal for int() with base 10

I am writing an ATM code and I am having trouble with a bit of code, which despite my best efforts I cannot fix. Here is the error message:

line 35, in Withdraw
back2_menu = int(input("Do You Wish To Have Another Transation? "))
ValueError: invalid literal for int() with base 10: 'Y'

And here is the troublesome piece of code

def Withdraw2(self):
    amount = int(input("How Much Do You Wish To Withdraw: \n £"))
    if int(amount) <= self.balance:
        self.balance = self.balance - amount
        print("Withdrawl Accepted. \nYour New Balance Is: £" + str(self.balance))
    else:
        print("Withdrawl Denied. \n You Have £" + str(self.balance) + "Within Your Account")
        again = int(input("Do You Wish To Enter Another Amount"))
        if again in ("Y", "y", "Ye", "ye", "Yes", "yes"):
            print(atm.Withdraw2())
        else:
            backmenu = str(input("Do You Wish To Have Another Transation? "))
            if backmenu in ("Y", "y", "Ye", "ye", "Yes", "yes"):
                print(atm.Menu())
            else:
                print(atm.End())

I am just wondering how to fix this so my program runs smoothly. Many Thanks

Upvotes: 1

Views: 2740

Answers (1)

David Elson
David Elson

Reputation: 221

The int(...) function turns a string integer (i.e. "123") into an integer integer (i.e. 123) without quotes.

"y" does not conform to integer syntax, so int("y") is not valid.

Instead of:

again = int(input("Do You Wish To Enter Another Amount"))

Do something like a few lines further, and say:

again = str(input("Do You Wish To Enter Another Amount"))

Then you won't get that error, which is caused by trying to parse out an integer from 'Y'.

Note that the earlier input:

amount = int(input("How Much Do You Wish To Withdraw: \n £"))

Is valid, if you enter a number. Beware, though, that int(3.4) will truncate to 3.

Upvotes: 1

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