石头剪刀布python程序输入错误答案

【问题标题】：rock paper scissor python program input error石头剪刀布python程序输入错误
【发布时间】：2022-02-13 17:09:09
【问题描述】：

随机导入

def rpc(): a = [“石头”、“纸”、“剪刀”]

b = random.choice(a)

print(b)

userinput = input('please type rock, paper or scissor:')
if userinput == "":
    print("please print the right thing")
    rpc()

if userinput != "rock":
    print("type the right thing!")
    rpc()

if userinput != "paper":
    print("type the right thing!")
    rpc()

if userinput != "scissor":
    print("type the right thing!")
    rpc()


while b == userinput:
    print("it is a draw, do you want to play another one?")
    c = input("type 'y' if you want to play one more time or press 'n': ")
    if c == 'y':
        rpc()
    elif c =='n':
        break
    else:
        print("print the right words please")

if b == 'rock' and userinput == 'paper':
    print("you win!")
    rpc()

elif b == 'rock' and userinput == 'scissor':
    print("you lost")
    rpc()

elif b == 'paper' and userinput == 'scissor':
    print("you win!")
    rpc()

elif b == 'paper' and userinput == 'rock':
    print("you lost!")
    rpc()

elif b == 'scissor' and userinput == 'rock':
    print("you win!")
    rpc()

elif b == 'scissor' and userinput == 'paper':
    print("you lost!")
    rpc()

rpc()

这是我的石头剪刀布代码，它非常简单，但是当我运行我的代码并输入石头剪刀布时，我得到了请打印正确的东西声明，我不知道它为什么会发生，任何帮助太好了，谢谢！

【问题讨论】：

不直接相关：适当地命名您的变量。这可能会花费您几毫秒的打字时间，但它会节省您思考代码的时间。
并将您的代码拆分为逻辑片段。在这里你可以有一个game() 函数、一个ask_move() 函数、一个compute_winer() 函数和一个show_result() 函数。虚构的例子，但你明白了重点：每个功能都应该专注于一个容易推理的事情。
看这里stackoverflow.com/questions/33190894/rock-paper-scissors-game
Rock-Paper-Scissors Game的可能重复
当您修复所有其他问题时，您会遇到最大深度递归错误：一个了解此网站名称含义的机会

标签： python

【解决方案1】：

让我们清理一下......

userinput = input('please type rock, paper or scissor:')
while userinput not in acceptable_inputs:
    userinput = input('please type rock, paper or scissor:')
opponents_choice = random.choice(objects)
# Check and print. Loop to keep playing - 2 out of 3 wins....

【讨论】：

【解决方案2】：

import random
def winner(riya,siya):
    if (riya==siya):
        return None
    elif (riya=="s"):
        if (siya=="p"):
            return False
        elif(siya=="k"):
            return True
    elif (riya=="p"):
        if (siya=="s"):
            return False
        elif (siya=="k"):
            return True
    elif (riya=="k"):
        if (siya=="s"):
            return True
        elif (siya=="p"):
            return False
        
print("riya pick: stone(s) paper(p) sessior(k)") 
a= random.randint(1,3) 
if a==1:
    riya="s"
elif a==2:
    riya="p" 
elif a==3:
    riya="k"
siya=input(" siya pick: stone(s) paper(p) sessior(k)")
b=winner(riya,siya)
print(f"riya pick{riya}")
print(f"siya pick {siya}")
if b==None:
    print("tie")
elif b:
    print("winner")`enter code here`
else:enter code here
    print("loose")

【讨论】：