这个python程序应该检查匹配的字符串，但在三次错误输入之后，编程并没有终止

//匹配字符串程序

exact_word = "Guessing"
    guess_kid = ""
    guess_count = 0
    guess_limit = 3
    out_of_guesses = False

    while guess_kid != exact_word.casefold() and not(out_of_guesses):
        if guess_count < guess_limit:
            guess_kid = input("Enter another word : ")
            guess_count += 1
    else:
        out_of_guesses = True
    if out_of_guesses:
        print("Out of luck")
    else:
        print("you win")

---

//三次错误输入后程序未终止

缩进触发。试试这个：

exact_word = "Guessing"
guess_kid = ""
guess_count = 0
guess_limit = 3
out_of_guesses = False

while guess_kid != exact_word.casefold() and not(out_of_guesses):
    if guess_count < guess_limit:
        guess_kid = input("Enter another word : ")
        guess_count += 1
    else:
        out_of_guesses = True

        if out_of_guesses:
            print("Out of luck")
        else:
            print("you win")

在使用循环/子句之后，跟踪缩进是很重要的。因为我们在while:loop之前没有缩进，所以我将前5行移到完全左对齐

因为while:loop启动一个循环，所以下一个if子句和相应的else子句一个缩进。这些下的子集文本将再次缩进

希望这有帮助

exact_word = "Guessing"
guess_kid = ""
guess_count = 0
guess_limit = 3
out_of_guesses = False

while guess_kid != exact_word.casefold() and not(out_of_guesses):
    if guess_count < guess_limit:
        guess_kid = input("Enter another word : ")
        guess_count += 1
    else:
        out_of_guesses = True

if out_of_guesses:
        print("Out of luck")
else:
        print("you win")

这对你有用。

---

一旦在猜测之外进行了一些更改，您需要检查条件。

缩进关闭。第一个else子句应与while循环后设置的内部if子句OUT_of_猜测匹配，因为else块引用的是while，而不是while块中的if。一次输入后，程序将终止。否则：你会看到的。