Question

我正在尝试创建一个简单的轮盘游戏，作为python的初学者，我遇到了一些问题。

到目前为止，我有下面的代码，并创建了一些函数，试图让用户输入激活不同的功能。我已经在红色/黑色和奇数/偶数函数中创建了一些测试语句，但是，无论输入是什么，我只获得红色/黑色函数的输出（如果输入红色则为10，对于任何其他输入则为20）。任何帮助表示赞赏！

import random, time


# the spin and result of the roulette wheel
def spin():
    print("Wheel is spinning...")
    time.sleep(3)
    land_number = random.randint(0,36)
    print (land_number)

def number_chosen(x):
    pass

def red_black_chosen(chosen_bet):
if chosen_bet == 'red':
    red = 10
    print(red)
else:
    black = 20
    print(black)

def odd_even_chosen(chosen_bet):
   if chosen_bet == 'odd':
       odd = 'qwe'
       print(odd)
   else:
       even = 'abc'
       print(even)

def high_low_chosen(chosen_bet):
    pass

def dozen_chosen(chosen_bet):
    pass

#choice of bet
def betters_choice():
    print("""How would you like to bet? \n
             Choose a number
             Choose red or black
             Choose odd or even
             Choose low(1-18) or high(19-36
             Choose 1st dozen(1-12), 2nd dozen(13-24) or 3rd dozen(25-
36)""")
    global chosen_bet
    chosen_bet = input()
    if type(chosen_bet) == int:
        if chosen_bet < 0:
            print("Bet must be between 0 and 36")
        elif chosen_bet > 36:
            print("Bet must be between 0 and 36")
        else:
           number_chosen(chosen_bet)
    elif chosen_bet == 'red' or 'black':
        red_black_chosen(chosen_bet)
    elif chosen_bet == 'odd' or 'even':        
        odd_even_chosen(chosen_bet)
    elif chosen_bet == 'low' or 'high':
        high_low_chosen(chosen_bet)
    elif chosen_bet in ['1st dozen', '2nd dozen', '3rd dozen']:
        dozen_chosen(chosen_bet)
    else:
        print("Incorrect bet chosen")
        betters_choice()

betters_choice()

Answer 1

这是因为if函数不接受这样的多个条件。它们应该单独说明：

import random, time


# the spin and result of the roulette wheel
def spin():
    print("Wheel is spinning...")
    time.sleep(3)
    land_number = random.randint(0,36)
    print (land_number)

def number_chosen(x):
    pass

def red_black_chosen(chosen_bet):
    if chosen_bet == 'red':
        red = 10
        print(red)
    else:
        black = 20
        print(black)

def odd_even_chosen(chosen_bet):
    if chosen_bet == 'odd':
        odd = 'qwe'
        print(odd)
    else:
        even = 'abc'
        print(even)

def high_low_chosen(chosen_bet):
    pass

def dozen_chosen(chosen_bet):
    pass

#choice of bet
def betters_choice():
    print("""How would you like to bet? \n
             Choose a number
             Choose red or black
             Choose odd or even
             Choose low(1-18) or high(19-36
             Choose 1st dozen(1-12), 2nd dozen(13-24) or 3rd dozen(25-
36)""")
    global chosen_bet
    chosen_bet = input()
    if type(chosen_bet) == int:
        if chosen_bet < 0:
            print("Bet must be between 0 and 36")
        elif chosen_bet > 36:
            print("Bet must be between 0 and 36")
        else:
            number_chosen(chosen_bet)
    elif chosen_bet == 'red' or chosen_bet =='black':
        red_black_chosen(chosen_bet)
    elif chosen_bet == 'odd' or chosen_bet =='even':        
        odd_even_chosen(chosen_bet)
    elif chosen_bet == 'low' or chosen_bet =='high':
        high_low_chosen(chosen_bet)
    elif chosen_bet in ['1st dozen', '2nd dozen', '3rd dozen']:
        dozen_chosen(chosen_bet)
    else:
        print("Incorrect bet chosen")
        betters_choice()

betters_choice()

Answer 2

addEventListener('load')始终属于chosen_bet类型，因为input始终返回字符串。

首先尝试在str上致电isdigit()。

chosen_bet

我还修复了if chosen_bet.isdigit(): # then check values and such if int(chosen_bet) < 0: elif chosen_bet == 'red' or chosen_bet == 'black' # rest of your code声明中的错误。您需要对elif两边的chosen_bet进行比较。

功能选择问题

2 个答案: