代码跳过第一个 IF 语句，有没有办法修复[重复]

Question

好的，原来的问题得到了解答（多行代码循环和代码无法识别，或者，如果您需要帮助，请在下面查找这些答案），但这是一个新问题：我的游戏无法识别正确的数字 这是 v3 代码：

while True:
answer = randint(1,10)
input1 = input('choose a number between 1 and 10.')
if answer == input1:
    print('you got it, good job.')
    input2 = input('do you want to play again?')
    if input2 != 'yes' and input2 != 'ys' and input2 != 'y' and input2 != 'ye' and input2 != 'yeah' and input2 != 'yup' and input2 != 'sure' and input2 != 'alright':
        print("ok, stopping")
        break
else:
    print ("wrong answer, the answer was")
    print(answer)
    input2 = input('do you want to play again?')
    if input2 != 'yes' and input2 != 'ys' and input2 != 'y' and input2 != 'ye' and input2 != 'yeah' and input2 != 'yup' and input2 != 'sure' and input2 != 'alright':
        print("ok, stopping")
        break

这次是游戏部分的问题:L, 仍然使用 python 3.12.3 IDLE。 这是问题的图片： 我得到了正确的答案，代码仍然认为是错误的？！？！？

我重复使用该帖子的原因是我直到明天才能提出另一个问题

Answer 1

要继续执行某些操作直到用户想要停止，请将整个代码包装在

while True:

循环中，并在用户准备退出时跳出循环。

while True:

    # game code goes here

    answer = input("Do you want to play again?")
    if answer != "yes":
        break

Answer 2

您的 if 语句的条件逻辑有缺陷。因为您使用 or 检查响应是否不等于，所以它始终为真。 （F 或 T 或 T 或 T 始终为真）。

正确的条件逻辑是检查输入是否不等于all响应 - 因此使用and而不是or。

while True:
input1 = input('choose a number between 1 and 10.')
if input1 == answer:
    print('you got it, good job.')
    input2 = input('do you want to play again?')
    if input2 != 'yes' and input2 != 'ys' and input2 != 'y' and input2 != 'ye' and input2 != 'yeah' and input2 != 'yup' and input2 != 'sure' and input2 != 'alright':
        print("ok, stopping")
        break
elif input1 != answer:
    print ("wrong answer, the answer was")
    print(answer)
    input2 = input('do you want to play again?')
    if input2 != 'yes' and input2 != 'ys' and input2 != 'y' and input2 != 'ye' and input2 != 'yeah' and input2 != 'yup' and input2 != 'sure' and input2 != 'alright':
        print("ok, stopping")
        break

除了您编写的内容之外，您还可以将代码编辑为：

input_continue = ['yes', 'ys', 'y', 'ye', 'yeah', 'yup', 'sure', 'alright']

while True:
input1 = input('choose a number between 1 and 10.')
if input1 == answer:
    print('you got it, good job.')
    input2 = input('do you want to play again?')
    if input2 not in input_continue:
        print("ok, stopping")
        break
elif input1 != answer:
    print ("wrong answer, the answer was")
    print(answer)
    input2 = input('do you want to play again?')
    if input2 not in input_continue:
        print("ok, stopping")
        break

这将 1) 产生所需的条件检查，2) 更易于阅读。