Python相等运算符==条件判断异常：输入任意值均返回True

我做了个文本小游戏，想加个重新玩的选项，就把整个游戏逻辑套进了循环，然后在循环末尾加了这段代码：

again = input('Would you like to play again? (Y/N) ')
if again == "Y" or "y":
    continue
else:
    break

结果不管输入啥，这个条件都返回True。我改了代码调试：

again = input('Would you like to play again? (Y/N) ')
print(again)
if again == "Y" or "y":
    print("True")
else:
    print("False")

输入n的时候输出是这样的：

Would you like to play again? (Y/N) n
n
True

为啥会这样？ #

你写的if again == "Y" or "y"在Python里的实际逻辑是(again == "Y") or "y"。Python里非空字符串属于“真值”，所以不管前面的again == "Y"是对是错，整个表达式都会因为后面的"y"永远返回True。

怎么改？ #

有几种简单的修复方式：

1. 分别判断两个相等条件：

again = input('Would you like to play again? (Y/N) ')
if again == "Y" or again == "y":
    continue
else:
    break

2. 把输入转成小写（或大写）后再判断，简化代码：

again = input('Would you like to play again? (Y/N) ').lower()
if again == "y":
    continue
else:
    break

3. 用in判断输入是否在指定的选项集合里：

again = input('Would you like to play again? (Y/N) ')
if again in ("Y", "y"):
    continue
else:
    break

这样就能正确根据输入决定是否重新开始游戏了。

内容的提问来源于stack exchange，提问作者Michael Mei