How do you check if a input value is int and otherwise put in a loop?

Question

I'm a very beginner in computer language (2 days) and am writing a basic slot machine code in python.

import random
import time

balance = 1000

print("Welcome to the Slot World")

player_name = input("What's your name?")


while balance > 0:

    print(player_name, ", you have $",balance, " to bet.")
    bet = int(input("How much do you want to bet?"))

    x = random.randint(1, 5)
    y = random.randint(1, 5)
    z = random.randint(1, 5)
    numbers = [x, y, z]

    if bet != int:
        print("Please enter correct amount")
        continue
    elif bet > balance:
        print("You do not have sufficient amount")
        continue
    elif bet <= 0:
        print("Please enter correct amount")
        continue
    else:
        print("You have bet $", bet)
        time.sleep(0.5)
        print(numbers)
        balance = balance - bet

    if x == y and y == z:
        if x == 1:
            print("Winner 20x!")
            bet = bet * 20
            balance = balance + bet
        elif x == 2:
            print("Winner 30x!")
            bet = bet * 30
            balance = balance + bet
        elif x == 3:
            print("Winner 40x!")
            bet = bet * 40
            balance = balance + bet
        elif x == 4:
            print("Winner 50x!")
            bet = bet * 50
            balance = balance + bet
        elif x == 5:
            print("Winner 100x!")
            bet = bet * 100
            balance = balance + bet

    elif x == y and not y == z:
        if x == 1:
            print("Winner 2x")
            bet = bet * 2
            balance = balance + bet
        elif x == 2:
            print("Winner 3x")
            bet = bet * 3
            balance = balance + bet
        elif x == 3:
            print("Winner 4x")
            bet = bet * 4
            balance = balance + bet
        elif x == 4:
            print("Winner 5x")
            bet = bet * 5
            balance = balance + bet
        elif x == 5:
            print("Winner 10x")
            bet = bet * 10
            balance = balance + bet

    elif not x == y and y == z:
        if y == 1:
            print("Winner 2x")
            bet = bet * 2
            balance = balance + bet
        elif y == 2:
            print("Winner 3x")
            bet = bet * 3
            balance = balance + bet
        elif y == 3:
            print("Winner 4x")
            bet = bet * 4
            balance = balance + bet
        elif y == 4:
            print("Winner 5x")
            bet = bet * 5
            balance = balance + bet
        elif y == 5:
            print("Winner 10x")
            bet = bet * 10
            balance = balance + bet

    elif not x == y and x == z:
         if x == 1:
                print("Winner 2x")
                bet = bet * 2
                balance = balance + bet
         elif x == 2:
                print("Winner 3x")
                bet = bet * 3
                balance = balance + bet
         elif x == 3:
                print("Winner 4x")
                bet = bet * 4
                balance = balance + bet
         elif x == 4:
                print("Winner 5x")
                bet = bet * 5
                balance = balance + bet
         elif x == 5:
                print("Winner 10x")
                bet = bet * 10
                balance = balance + bet

    else:
        print("Bad Luck! You Lost!")
print("You lost all your money\nThe End")

If I input a character for bet

bet = int(input("How much do you want to bet?"))

it gives me

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

what is your desired behavior? ask for correct input until it is not convertible to `int`? — Azat Ibrakov, Aug 16 '18 at 06:17

Joe · Answer 1 · 2018-08-16T06:28:27.480

1

You can use try:

while True:
    try:
        bet = int(input("How much do you want to bet?"))
    except ValueError:
        print "Not valid integer"
        continue
    break

edited Aug 16 '18 at 06:28

answered Aug 16 '18 at 06:14

Joe

12,057
5
39
55

1

using `except` without specifying exception type is a bad thing to do – Azat Ibrakov Aug 16 '18 at 06:15
@Joe, add the specific Exception that is to be caught. In this case `ValueError` – FlyingTeller Aug 16 '18 at 06:27

utks009 · Answer 2 · 2018-08-16T06:20:52.560

1

You can not change string like a-Z into integer so you have to check for it

try:
    bet = int(input("How much do you want to bet?"))
except ValueError:
    print ("Not valid integer")
    continue

edited Aug 16 '18 at 06:20

answered Aug 16 '18 at 06:16

utks009

573
4
14

Hein Wessels · Answer 3 · 2018-08-16T06:20:00.787

0

Firstly, the letter a can't be converted to a number, because what to which number should it then convert?

So for that @Joe's answer with try try except should be perfect.

Also, it might be safer to rather convert to a float. It makes it more robust, you can enter decimal values if you want,

bet = float(input("How much do you want to bet?"))

edited Aug 16 '18 at 06:20

answered Aug 16 '18 at 06:16

Hein Wessels

937
5
15

If you enter the char `a` does your solution solve the problem? :) – Joe Aug 16 '18 at 06:17
Just realized >.< I'm updating my answer. – Hein Wessels Aug 16 '18 at 06:18

score 0 · Answer 4 · answered Aug 16 '18 at 06:18

Use int() function to try to convert string value to integer. It will raise ValueError if it is not. Use try to catch it and use its else clause to break from the loop.

while True:
    try:
        value = int(input('Enter your number'))
    except ValueError:
        pass
    else:
        break

How do you check if a input value is int and otherwise put in a loop?

4 Answers4