python 剪刀石头布

Question

这个问题已经有答案了:

Why does "a == x or y or z" always evaluate to True? How can I compare "a" to all of those? (8 个回答)

已关闭 7 年前。

我是编程新手，正在尝试使用 python 2.7 构建一个简单的石头剪刀布程序。在我的函数中，我有 2 个主要的 if 语句

rules = raw_input("Before we being playing would you like to hear the rules first? ")
if rules.lower() == "yes":
    print """ 
    Scissors cuts Paper 
    Paper covers Rock 
    Rock crushes Scissors"""

第二个

choice = raw_input("please enter your choice? (Must be either rock, paper or scissors) ")
computer = random.choice(["rock", "paper", "scissors"])

if choice == "rock" or "paper" or "scissors": 
    if choice == computer : 
        print "Its a tie !"
    elif choice == "rock" and computer == "scissors": 
        print "Rock beats scissors you win!."

    elif choice == "rock" and computer == "paper": 
        print "Paper beats rock you loose !."

    elif choice == "paper" and computer == "scissors": 
        print "Scissors beats paper you loose !."

    elif choice == "paper" and computer == "rock": 
        print "Paper beats rock you win !."

    elif choice == "scissors" and computer == "paper": 
        print "Scissors beats paper you win !."

    elif choice == "scissors" and computer == "rock": 
        print "Rock beats scissors you loose !."
else : 
    print "Invalid Entry Please try again."

这两段代码分别按其应有的方式工作，但是当我尝试将它们放在一个函数中时，第一个 if 语句询问规则有效，但随后在具有程序主要功能的第二个 if 语句之前退出。我尝试在第一个 if 语句中缩进第二位代码，但似乎不起作用

我想知道是否有办法让这两个代码片段在一个简单的函数中工作？或者我应该创建一个具有这两个函数的类？另外，如果有人对如何使我的程序变得更好有任何建议，请告诉我。感谢您提前提供的任何帮助。

这是完整代码

import random 

def rock_paper_scissors_spock():

    rules = raw_input("Before we being playing would you like to hear the rules first? ")
    if rules.lower() == "yes":
        print """ 
        Scissors cuts Paper 
        Paper covers Rock 
        Rock crushes Scissors"""

    choice = raw_input("please enter your choice? (Must be either rock, paper or  scissors) ")
    computer = random.choice(["rock", "paper", "scissors"])

    if choice == "rock" or "paper" or "scissors": 
        if choice == computer : 
            print "Its a tie !":

        elif choice == "rock" and computer == "scissors": 
            print "Rock beats scissors you win!."

        elif choice == "rock" and computer == "paper": 
            print "Paper beats rock you loose !."

        elif choice == "paper" and computer == "scissors": 
            print "Scissors beats paper you loose !."

        elif choice == "paper" and computer == "rock": 
            print "Paper beats rock you win !."

        elif choice == "scissors" and computer == "paper": 
            print "Scissors beats paper you win !."

        elif choice == "scissors" and computer == "rock": 
            print "Rock beats scissors you loose !."
    else : 
        print "Invalid Entry PLease try again." 


rock_paper_scissors_spock()

Answer 1

你说if choice == "rock"or "paper"or "scissors":，但是Python并没有将choice ==连接到所有的选择。你可以在 choice == "rock" 两边加上括号，它会做同样的事情。将其更改为 if choice in ("rock", "paper", "scissors")