在“if”条件下使用变量时出现Python3 NameError

Question

我正在尝试为“石头剪纸”编写一个简单的程序，它只是一个初级代码，但是我面临着一个变量的NameError，这个变量是我定义的，并且使用这个变量作为if的条件。我使用的是Python3。我在这里发布了我编写的程序：

player1count = 0
player2count = 0
print("Make a Choice: \n1.Stone\n2.Paper\n3.Scissor")
player1 : int(input("Player 1 plays "))
player2 : int(input("Player 2 plays "))
while (True):
    if player1 is 1 and player2 is 3 or player1 is 2 and player2 is 1 or player1 is 3 and player2 is 2:
        player1count += 1
        print(player1+" beats "+player2)
    elif player2 == 1 and player1 == 3 or player2 == 2 and player1 == 1 or player2 == 3 and player1 == 2:
        player2count += 1
        print(player2 + " beats " + player1)
    else:
        continue
    if(player1count == 5):
        print("Player 1 Wins!")
        break
    elif(player2count == 5):
        print("Player 2 Wins!")
    else:
        continue

这是我得到的错误：

Make a Choice: 
1.Stone
2.Paper
3.Scissor
Player 1 plays 1
Player 2 plays 3
Traceback (most recent call last):
  File "G:/Study/Python/StonePaperScissor.py", line 7, in <module>
    if player1 is 1 and player2 is 3 or player1 is 2 and player2 is 1 or player1 is 3 and player2 is 2:
NameError: name 'player1' is not defined

Process finished with exit code 1

Answer 1

player: type只是一个类型提示。它实际上不做任何事情，它是用来帮助代码分析的可读性和外部工具。它也不会像静态语言那样声明变量。你知道吗

在python中，要声明变量，必须给它赋值。所以呢

player1: int = int(input("Player 1 plays "))

使用类型提示和值赋值。你知道吗

注意：您想要的是player1==1，而不是player1 is 1。is检查它们是否是同一个对象，而不是是否相等。你知道吗