Python如果语句不起作用
问题说明
我有一段无效的代码。
if difficulty=="Easy" or "easy" or "1":
with open("EasyQs.csv") as f:
allData = [line.strip().split(",") for line in f]
questions = [data[0] for data in allData]
answers = [data[1] for data in allData]
print(questions)
if difficulty=="Hard" or "hard" or "2":
with open("MediumQs.csv") as f:
allData = [line.strip().split(",") for line in f]
questions = [data[0] for data in allData]
answers = [data[1] for data in allData]
print(questions)
我尝试输入2，hard或Hard，但是它总是从 easy CSV中打印问题？
为什么？以及如何解决？
I try to type 2, hard or Hard but it always prints the questions from the 'easy' CSV? Why is this? and how can it be solved?
正确答案
在Python中，如果您要检查某项是否是一组值之一，您可以使用以下方法：
In Python, if you want to check if something is one of a set of values, you use this:
if difficulty in ("Easy", "easy", "1"):
原因是您当前正在执行的操作不起作用就像您认为的那样。您具有的条件：
The reason is that what you're doing right now doesn't work like you think it does. The conditional you have:
if difficulty == "Easy" or "easy" or "1":
实际评估如下：
if (difficulty == "Easy") or ("easy") or "1":
因为或是比 == 宽松的绑定运算符。因此，此总体合并条件将始终为true，因为 easy 是真实值，因此即使（difficulty == Easy）为假，或运算符会将其右侧评估为true并将其返回。
Because or is a looser binding operator than ==. So this overall combined condition will always be true, because "easy" is a true value, so even if (difficulty == "Easy") is false, the or operator will evaluate its right hand side to true and return it.
这就是为什么现在您的简单案例总是会触发。
That's why right now your "easy" case always triggers.
这篇好文章是转载于：学新通技术网
