我必须制作一个石头剪刀布游戏，但是当我尝试将输入转换为字符串时，if 语句不起作用 str(.....)
我在这里使用整数来确保代码正常工作，否则它就不能处理字符串。
当我运行此代码时，我也无法将输入加下划线，如何做到这一点？
player_1 = str(input("Enter Player 1 choice (R, P, or S): "))
player_2 = str(input("Enter Player 2 choice (R, P, or S): "))
if player_1 == S and player_2 == S:
print("A tie!")
elif player_1 == R and player_2 == R:
print("A tie!")
elif player_1 == P and player_2 == P:
print("A tie!")
elif player_1 == R and player_2 == 2:
print("Rock beats scissors! Player 1 wins.")
elif player_1 == S and player_2 == R:
print("Rock beats scissors! Player 2 wins.")
elif player_1 == 9 and player_2 == R:
print("Paper beats rock! Player 1 wins.")
elif player_1 == R and player_2 == P:
print("Paper beats rock! Player 2 wins.")
elif player_1 == S and player_2 == P:
print("Scissors beat paper! player 1 wins.")
elif player_1 == P and player_2 == S:
print("Scissors beat paper! player 2 wins.")
每次运行代码时都会出现此错误：
Enter Player 1 choice (R, P, or S): S
Enter Player 2 choice (R, P, or S): S
Traceback (most recent call last):
File "D:\CP 104\**********\src\t03.py", line 16, in <module>
if player_1 == S and player_2 == S:
NameError: name 'S' is not defined
我究竟做错了什么？
对于您的每个案例，您都在根据字符串检查变量的值。每个字符串的字符串可以是“R”、“S”或“P”，在你的 if 语句中，你应该这样写
例如
if player_1 == "S" and player_2 == "S":
等等。
正如@jedwards 所说，您正在使用 python3，并且 input() 返回一个字符串，因此不需要第一行和第二行中的 str() 包装器
你可以简单地说
player_1 = input("Enter Player 1 choice (R, P, or S): ")
player_2 = input("Enter Player 1 choice (R, P, or S): ")
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