或 & 和 if 语句在一行中没有按预期工作

Question

我不确定为什么这段代码没有按预期工作，有人可以帮我吗？

# check if a user eligible for watching coco movie or not.
# if user name starts with "a" or "A" and his age is greater then 10 then he can, else not.

user_name = input("please enter your name in letters here: ")
user_age = int(input("please enter your age in whole numbers here: "))

if user_name[0] == ("a" or "A") and user_age > 10 :
  print("you can watch coco movie")
else:
  print("sorry, you cannot watch coco")

我在所有可能的条件下测试了这段代码，它工作正常，但最后一个条件没有按预期工作，最后一个条件不知道为什么条件为 False。

我在这里粘贴所有测试条件和 IDLE 的结果：

please enter your name in letters here: a
please enter your age in whole numbers here: 9
sorry, you cannot watch coco
>>> 
====================== RESTART: D:\MyPythonScripts\1.py ======================
please enter your name in letters here: a
please enter your age in whole numbers here: 10
sorry, you cannot watch coco
>>> 
====================== RESTART: D:\MyPythonScripts\1.py ======================
please enter your name in letters here: a
please enter your age in whole numbers here: 11
you can watch coco movie
>>> 
====================== RESTART: D:\MyPythonScripts\1.py ======================
please enter your name in letters here: A
please enter your age in whole numbers here: 9
sorry, you cannot watch coco
>>> 
====================== RESTART: D:\MyPythonScripts\1.py ======================
please enter your name in letters here: A
please enter your age in whole numbers here: 10
sorry, you cannot watch coco
>>> 
====================== RESTART: D:\MyPythonScripts\1.py ======================
please enter your name in letters here: A
please enter your age in whole numbers here: 11
sorry, you cannot watch coco
>>> 

Answer 1

在您编写的代码中，user_name[0] 与表达式（'a' 或 'A'）进行比较。表达式（'a' 或 'A'）的计算结果为 'a'。试试这个，

print( 'a' or 'A' )

结果是

a

因此，仅当 user_name 以 'a' 开头且年龄大于 10 时，条件才测试为真。

这是一个代码 sn-p，它可以满足您的预期：

if user_name[0] in 'aA' and user_age > 10 :
  print("you can watch coco movie")
else:
  print("sorry, you cannot watch coco")

Answer 2

上一个测试用例没有得到预期值的原因是这种情况

user_name[0] == ("a" or "A")

您看到（“a”或“A”）的评估方式与您想象的不同。括号使它成为它自己的表达式。 该表达式基本上是说 If 'a' is null then return 'A'

因此，这总是返回 'a'，返回那个输出。

user_name[0] == "a" or user_name[0] == "A"

应该解决这个问题

查看这篇文章以获得更多解释 https://stackoverflow.com/a/13710667/9310329

干杯