或＆和if语句在一行中无法按预期工作

Question

我不确定为什么该代码无法按预期工作，请问有人可以帮我吗？

# 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此代码，它可以正常工作，但最后一个条件未按预期工作，在最后一个条件下，不知道为什么条件为False。

我在这里粘贴所有经过测试的条件，并且从IDLE获得结果：

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，将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”。 尝试这个，

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true。

这是一个代码片段，它可以实现您的预​​期目的：

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

您的最后一个测试用例没有获得期望值的原因是这种情况

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

您会看到（“ a”或“ A”）的评估结果与您想像的有所不同。 括号使它成为自己的表达。 该表达式基本上说如果'a'为null，则返回'A'

因此，它始终返回“ a”，并返回该输出。

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