I'm getting invalid literal for int() with base 10: on this code

Question

on this Code:

SERVICE_CHARGE = 2 TICKET_PRICE = 10 ticketsRemaining = 100

def calculatePrice(numberOfPeople): return (numberOfPeople*TICKET_PRICE)+SERVICE_CHARGE

while ticketsRemaining >=1: print("Tickets Remaining {}".format(ticketsRemaining))

userName = (input("What is your Name?  "))
numberOfPeople = input("How many tickets would you like, {}?  ".format(userName))

try: 
    numberOfPeople = int (numberOfPeople)
    if numberOfPeople > ticketsRemaining:
        raise ValueError ("There are only {} tickets remaining".format(ticketsRemaining))

except ValueError as err: 
    print("Oh No, That's not a valid input, {}.".format(err))

else:
    amountDue= calculatePrice(numberOfPeople)
    print("Total Cost ${}" .format(amountDue))
    shouldProcced = input("Do you want to proceed ? (yes/no)    ")

    if shouldProcced.lower() == "yes":
        print("SOLD!!")
        ticketsRemaining = (ticketsRemaining - numberOfPeople)
    else:
        print("Thank You, {}".format(userName))

print("Sorry the tickets are all sold out")

Answer 1 · 2020-07-11T01:34:23Z

July 11, 2020 1:34am

The error you are getting invalid literal for int() with base 10: is because you are passing in a string argument which can't be converted to an int, also the message invalid literal for int() with base 10: is the systems' way of handling this specific error, it might look that your code is incorrect or the error is weird but that is how the system handles this error.

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Mariel Vidal

Mariel Vidal

I'm getting invalid literal for int() with base 10: on this code

1 Answer

boi

boi