Question

嘿伙计们我已经编写了下面的代码，一切似乎工作得很好，除非它应该是一个平局我一直得到“name2胜利”以及答案应该是无效的。我不知道出了什么问题，因为我的逻辑似乎是正确的。

print("Welcome to Rock, Paper, Scissors!")
print("Let's Begin ...")
name1 = input("Player 1: What's your name?")
name2 = input("Player 2: What's your name?")

print("Hello " + name1 + " and " + name2)
print(name2 + ": Close your eyes!")

choice1 = input(name1 + ": enter 'r' for rock, 'p' for paper, and 's' for Scissors: ")

print("Great choice! Now - cover your answer and ask " + name2 + " to choose. \n\n\n")
choice2 = input(name2 + ": enter 'r' for rock, 'p' for paper, and 's' for scissors: ")

if(choice1 == "r" , choice2 == "p"):
  print(name2 + " wins!")
elif(choice1 == "r" , choice2 == "s"):
  print(name1 + " wins!")
elif(choice1 == "r" , choice2 == "r"):
  print("It is a draw")
elif(choice2 == "r" , choice1 == "p"):
  print(name1 + " wins!")
elif(choice2 == "r" , choice1 == "s"):
  print(name2 + " wins!")
elif(choice1 == "p" , choice2 == "s"):
  print(name2 + " wins!")
elif(choice1 == "p" , choice2 == "p"):
  print("It is a draw!")
elif(choice2 == "p" , choice1 == "s"):
  print(name1 + " wins!")
elif(choice1 == "s" , choice2 == "s"):
  print("It is a draw!")
else:
  print("Invalid asnwer")


print("Thanks for playing Rock, Paper, scissors")

Answer 1

检查多个条件的整个方法不正确 -

if(choice1 == "r" , choice2 == "p"):

取得choice1 == "r", choice2 == "p"形成的元组的真实性，这将是真实的，因为它不会是空的，所以你永远不会达到elif陈述，而玩家2将永远胜利。

检查这两个条件的正确方法是使用and运算符，并且不需要括号。

if choice == "r" and choice2 == "p":

修复所有这些条件应该会有所帮助。

Answer 2

更改'和'的所有if条件中的'，'，如下所示：

if choice1 == "r" and choice2 == "p":
  print(name2 + " wins!")

注意：括号不是必需品。

为什么你总是得到“name2 wins”？

由于逗号，您在创建tuple的位置，如果不为空，则评估为true。

Answer 3

你可以做到


     choice = [choice1, choice2]
     if(choice == ['r', 'p']):
        print(name2 + " wins! ")
     elif(choice == ['r', 'p']):
        print(name1 + " wins!")
     elif(choice == ['r', 'r']):
        print("It is a draw")
     elif(choice == ['p', 'r']):
        print(name1 + " wins!")

我的“摇滚，纸，剪刀”代码有问题

3 个答案: