根据用户需要重复算法的次数(PYTHON 3.3.2 GUI)

Question

我试图重复这个算法，并在我的代码中使用值，请帮助!!

进口随机

def dice_roll(number):
    if number == 12:
        number = random.randint(1,12)
        print(number)
        return(number)
    elif number == 6:
         number = random.randint(1,12)
         print(number)
         return(number)
    else:
        number == 4
        number = random.randint(1,12)
        print(number)
        return(number)

dice_roll(int(input("你想掷哪个骰子? ")--> ")))

Answer 1

是这样的吗?

import random
def dice_roll(number):
    if number == 12:
        number = random.randint(1,12)
        print(number)
        return(number)
    elif number == 6:
        number = random.randint(1,12)
        print(number)
        return(number)
    else:
        number == 4
        number = random.randint(1,12)
        print(number)
        return(number)
while True:
   dice_roll(int(input("Which dice would you like to roll? --> ")))
   doRepeat=input('Go again?')
   if doRepeat == 'no':
      break

编辑:

 raw_input() was renamed to input().

你可以在3中使用"input()"。X，像旧版本一样，这个:

doRepeat=eval(input('Go again?'))

Answer 2

典型的方法是while True循环，当用户退出时，您将break，例如:

while True:
    user_input = int(input("Which die would you like to roll (0 to quit)? "))
    if user_input in (4, 6, 12):
        dice_roll(user_input)
    elif user_input:
        print("Not an acceptable die.")
    else:
        break

请注意，您的dice_roll函数目前总是掷12面骰子，并且可以简化:

def dice_roll(sides):
    number = random.randint(1, sides)
    print(number)
    return number

(这也取消了对4,6和12的限制，因此您可能需要相应地修改输入循环)。