如果陈述有效，为什么这不“不”？

Question

这是我的代码，

if diff != "1" or diff != "2" or diff != "3":
            print("You need to pick either 1, 2 or 3\n")

出于某种原因，结果是，

Pick a difficulty:
1) Easy
2) Medium
3) Hard
>> 2
You need to pick either 1, 2 or 3

我希望if语句检查变量 diff 是否等于字符串 1,2和3 。但是当我把 1,2或3 放入时，diff的错误信息不等于打印数字。为什么会这样？

Answer 1

要应用“not any”逻辑，您需要检查它是否是任何有效结果然后反转。 （NOR）

if not (diff == "1" or diff == "2" or diff == "3"):

或者应用DeMorgan's theorem这对于“不等于1而不等于2而不等于3”是公平的

if diff != "1" and diff != "2" and diff != "3":

当然python还有in和not in运算符，这使得它更加清晰：

if diff not in ("1", "2", "3"):

Answer 2

您需要使用而不是或。如果输入1，则diff != "1"将返回True。

您的代码应如下所示：

if diff != "1" and diff != "2" and diff != "3":
            print("You need to pick either 1, 2 or 3\n")