1

Okay so I basically want my code to work by asking the user for input on ROLL variable, if it's a number that is specified as a die (6,8,10,12,20) then that all works, it picks a random number and outputs it, easy. However I cannot make the loop quit, so it says while ROLL is not equal to 'quit', do the things. I'll post the code below, but idk how to make it work so that when I enter 'quit' it closes, right now when I do it it outputs the 'else' statement.

#!/usr/bin/env Python3

ROLL = ''

while ROLL != 'quit':

        import random
        ROLL=input("What sided Die would you like to roll(You may enter quit to close)? ")
        if ROLL == 6:
                print(random.randint(1, 6))
        elif ROLL == 8:
                print(random.randint(1, 8))
        elif ROLL == 10:
                print(random.randint(1,10))
        elif ROLL == 12:
                print(random.randint(1, 12))
        elif ROLL == 20:
                print(random.randint(1, 20))
        else:
                print("That's not an option please choose a die...")
print("Thank you")
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3
  • python 2 interprets your input. you cannot have an integer AND a string. python 3 would fail when comparing to numbers, python 2 fails when interpreting "quit" Commented Sep 16, 2018 at 20:01
  • I'm using 3 and to be honest I have no clue what you mean by that. I figure when I enter 'quit' the program would say "Okay, so ROLL is now quit, so I should kill the loop" Commented Sep 16, 2018 at 20:06
  • I should prolly remark that I'm just now learning to program in a real language. I'm teaching myself Python and Ruby. I taught myself how to use Linux and write simple stuff in Bash, but idk what your comment is supposed to tell me. Sorry I'm so newb lol Commented Sep 16, 2018 at 20:07

3 Answers 3

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1

I have tried to create a python 2 / 3 answer.

python 2 input interprets your input, converting to integers if possible for instance. So you cannot have an integer AND a string. When entering quit, since quit isn't known by the interpreter, you get a NameError

The only way to make it work would be to type "quit". But that'd be python 2 only still... Let's try to make it portable now.

python 3 fails when comparing strings to numbers, because input returns strings.

You'll have a hard time creating a code which works for both versions. I'd suggest this:

import random

# detect/create a portable raw_input for python 2/3
try:
    raw_input
except NameError:
    raw_input = input


while True:
    ROLL=raw_input("What sided Die would you like to roll(You may enter quit to close)? ")
    if ROLL.isdigit():
        ROLL = int(ROLL)

    if ROLL in [6,8,10,12,20]:
        print(random.randint(1, ROLL))
    elif ROLL == "quit":
        break
    else:
        print("That's not an option please choose a die...")

The first try/except block is there for python 2/3 compatibility. If raw_input exists, use it, else define it as input for python 3. From now on, raw_input is used, and returns strings not integers.

Now we have to add this: if ROLL.isdigit() tests if the string could be converted as an integer. If it's possible, it converts it.

Now we test if the reply is contained in the choice list, and if it is, we use ROLL for our random (avoids the ton of elif statements).

The loop has been turned to inconditional too, no need to initialize ROLL at start. Just break if quit is entered.

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3 Comments

Okay that makes more sense, thank you so much. I'm really loving the idea of designing my own programs and simple games, but I've found myself unduly angry a few too many times, I'm gonna check this out in a moment; I can have my computer at work but Football has us busy. I'll come back and let you know if it worked out.
Also thanks for showing me that shorthand array you did, instead of a bunch if/elif, the if ROLL in [array] is much more clean, I really appreciate that
THANKS! That worked out perfectly, I'll take note of some of the changes you made.
1

There's an easy fix. But first, the code, as posted above, does not work. Since ROLL is a string, the numbers in the if/elif statements need to be strings, too.

The solution (one easy solution) is to have an extra if statement for the "quit" case and then finish the while loop early using "continue", like so:

import random
ROLL = ''

while ROLL != 'quit':

    ROLL=input("What sided Die would you like to roll(You may enter quit to close)? ")
    if ROLL == 'quit':
        continue
    elif ROLL == '6':
        print(random.randint(1, 6))
    elif ROLL == '8':
        print(random.randint(1, 8))
    elif ROLL == '10':
        print(random.randint(1,10))
    elif ROLL == '12':
        print(random.randint(1, 12))
    elif ROLL == '20':
        print(random.randint(1, 20))
    else:
        print("That's not an option please choose a die...")

print("Thank you")

The "continue" will make the while loop start over, and thereby checking if ROLL is "quit". Since it is, it will terminate the loop and say "Thank you".

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0

The main error in your code is that input returns a string and you are comparing it to integers. By example '6' is not equal to 6.

Although, you could simplify your code greatly by comparing the input to a tuple of allowed value. This will be both shorter and more extendable.

from random import randint

while True:
    roll = input("What sided Die would you like to roll(You may enter quit to close)? ")

    if roll == 'quit':
        break
    elif roll in ('6', '8', '10', '12', '20'):
        print(randint(1, int(roll)))
    else:
        print("That's not an option please choose a die...")

print("Thank you")
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