Question

我是编码新手。我想尝试编写一个简单的剪刀石头布游戏。但是我不知道如何结束游戏。

在该程序的最后，如果用户输入错误，我想再次转到end变量。我尝试使用带注释的行，但不起作用。

TextField(
  controller:
      new TextEditingController(text: listDisplay[position].getBlockQty(),
  ),       
  textAlign: TextAlign.center,
  maxLines: 1,       
  keyboardType: TextInputType.numberWithOptions(
      decimal: true,
      signed: false),       
),

如果输入为“否”，我希望它结束游戏，如果输入为“是”，我希望它重新开始游戏，如果输入不正确，我希望再次出现提示。

Answer 1

只需检查end的值

if end is True:
    continue
else:
    break

由于您已经通过将input（）与“ yes”进行比较来将end的值设置为布尔值，所以它将说出用户是否要结束游戏？此外，您没有初始化输入变量，并且最后的elif条件将始终为true，如注释中所述。

Answer 2

您的代码有一些问题需要解决，并且可以简化一些地方。我对您的程序进行了一些更改，并添加了一些注释来解释这些更改。

player1 = input("What is player 1's name ? ").title() #This uses chaining to streamline code 
player2 = input("What is player 2's name ? ").title() #Same as above

while True:
    a = input(player1 + " What do you choose ? rock / paper / scissors : ") #no need to use a separate print statement
    b = input(player2 + " What do you choose ? rock / paper / scissors : ")

    valid_entries = ["rock", "paper", "scissors"] #To check for valid inputs

    if (a not in valid_entries) or (b not in valid_entries):
        print("Wrong input, try again")
        continue

    a_number = valid_entries.index(a) #Converting it to numbers for easier comparison
    b_number = valid_entries.index(b)

    if(a_number == b_number):
         print("Its a tie :-(")
    else:
        a_wins = ((a_number > b_number or (b_number == 2 and a_number == 0)) and not (a_number == 2 and b_number == 0)) #uses some number comparisons to see who wins instead of multiple if/elif checks

        if(a_wins):
            print(player1, "won !!!")
        else:
            print(player2, "won !!!")

    end = input("Do you want to play again ? yes/no ")

    while (end !="yes") and (end != "no"):
        print("invalid input, try again")
        end = input("Do you want to play again ? yes/no ")

    if end == "yes":
        continue
    else:
        print("GAME OVER")
        break

这些更改还通过使用另一个while循环进行检查，以查看重新启动游戏的输入是否有效

*请注意，我尚未测试这些更改，可能需要进行一些修改

Answer 3

好吧，您可以使用列表简化代码，然后简化if测试。您可以检查选项的顺序并根据其做出决定。您也可以使测试成为标准，以最大程度地减少if语句的数量。这是我改善您的代码的建议。希望对您有所帮助：

# get playe names
player1 = input("What is player 1's name ? ")
player2 = input("What is player 2's name ? ")
player1 = player1.title()
player2 = player2.title()

# init vars
options = ["rock", "paper", "scissors"]
players = [player1, player2]

# start game
while True:
    a = input(player1 + " What do you choose ? rock / paper / scissors : ")
    b = input(player2 + " What do you choose ? rock / paper / scissors : ")

    # check if inputs are correct
    while (a not in options or b not in options):
        print("Wrong input, Try again")
        a = input(player1 + " What do you choose ? rock / paper / scissors : ")
        b = input(player2 + " What do you choose ? rock / paper / scissors : ")

    # check who won
    if abs(options.index(a) - options.index(b)) == 1:
        print(players[1*int(options.index(a) > options.index(b))], "won !!!")

    elif abs(options.index(b) - options.index(a)) > 1:
        print(players[1*int(options.index(a) > options.index(b))], "won !!!")

    elif a == b:
        print("Its a tie :-(")

    # continue or drop game
    end = input("Do you want to play again ? yes/no ")
    if end == "yes":
        continue
    else:

        print('''

        GAME OVER''')
        break

如何在另一个while循环中使用if else语句

3 个答案: