首页 > 解决方案 > 如何修复 if 语句以使用字符串?

问题描述

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 语句将不起作用str(.....)

我在这里使用整数来确保代码正常工作,但它不适用于字符串。

当我运行此代码时,我也无法在输入下划线,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?

标签: pythonpython-3.xif-statementundefined-variable

解决方案


对于您的每种情况,您都在根据字符串检查变量的值。它们每个的字符串可以是“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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