有人可以解释一下这段代码是如何不正确的吗？

如何解决有人可以解释一下这段代码是如何不正确的吗？

我正在创建一段代码，可以将 24 小时时间转换为可以读取的时间（例如 1 分钟到午夜 12 点）。我已经快写完了代码，但是有一些bug，因为当分钟大于30时，我的代码会出错，例如2359会产生2359的时间是1分钟到晚上11点。这是我的代码

times = []
test_cases = int(input("How many test cases do you have?: "))

for _ in range(test_cases):
    user_input = input(str("What is your time? "))
    times.append(user_input)

for n in times:
    first_half = n[:len(n) // 2]
    second_half = int(n[len(n) // 2:])
    if int(first_half) == 12:
        first_half_2 = f"{12} noon"
    elif int(first_half) == 00:
        first_half_2 = f"{12} midnight"
    elif int(first_half) > 12:
        first_half_2 = f"{int(first_half) - 12}pm"
    else:
        first_half.strip("0")
        first_half_2 = f"{int(first_half)}am"

    if second_half == 00:
        second_half_2 = ""
    elif second_half == 1:
        second_half_2 = "a minute to "
    elif second_half == 15:
        second_half_2 = "a quarter past "
    elif second_half == 59:
        second_half_2 = "a minute to "
    elif (00 < second_half < 30) and second_half != 15:
        second_half_2 = f"{second_half} minutes past "
    elif second_half == 30:
        second_half_2 = "half past "
    elif (30 < second_half < 59) and second_half != 45:
        second_half_2 = f"{60 - second_half} minutes to "
    else:
        second_half_2 = "a quarter to "

    print(f"{n} is {second_half_2}{int(first_half_2) + 1}")

解决方法

我重写了一些，但尽量保留您的工作，以便于理解！首先查看分钟会有所帮助，因此您可以根据需要调整上半场。这不是最有效的方式，但应该比较容易理解。希望这会有所帮助：

while True:
    time = input("Enter the time or -1 quit: ")
    if time == "-1":
        break
    
    first_half = int(time[:len(time) // 2])
    second_half = int(time[len(time) // 2:])

    if first_half not in range(0,24) or second_half not in range(0,61):
        print("Not an actual time...")
        break
    
    if second_half == 0:
        second_half_2 = ""
    elif second_half == 15:
        second_half_2 = "a quarter past "
    elif second_half in range(1,30) and second_half!= 15:
        second_half_2 = f"{second_half} minutes past "
    elif second_half == 30:
        second_half_2 = "half past "
    elif second_half == 45:
        second_half_2 = "a quarter to " 
        first_half += 1
    elif second_half == 59:
        first_half += 1
        second_half_2 = "a minute to " 
    elif second_half in range(31,60) and second_half != 45:
        second_half_2 = f"{60 - second_half} minutes to "
        first_half += 1

    if first_half == 12:
        first_half_2 = f"{12} noon"
    elif first_half == 00 or first_half == 24:
        first_half_2 = f"{12} midnight"
    elif first_half in range(1,12):
        first_half_2 = f"{first_half}am"
    elif first_half in range(13,23):
        first_half_2 = f"{first_half - 12}pm"
    
    
    if len(str((first_half))) < 10:
        print(f"{time} is {second_half_2}{int(first_half_2[:1])}{first_half_2[1:]}")
    else:
        print(f"{time} is {second_half_2}{int(first_half_2[:2])}{first_half_2[2:]}")

对于初学者来说，测试用例输入是为了什么？是为了测试您的代码是否有效？

我重写了你代码的第一部分，假设测试用例不是必需的，因为我觉得没有它，程序可以做你想做的事。我希望这是一个公平的假设。这样做的作用是将军事时间分成几小时和几分钟，以便可以轻松地将它们分开处理，这似乎您已经知道该怎么做。

user_input = str(input("What is your time?"))
def split_num(n): return [str(i) for i in str(n)]
def recombine(lst): return ''.join(lst)

split_time = split_num(user_input)

hour = recombine(split_time[0:2])
minute = recombine(split_time[2:5])

至于您遇到的问题，我尝试修复它，但似乎用户输入集合存在其他一些更大的问题，这可能间接导致了该问题。我认为现在你可以用我上面写的代码更整齐地组织数据了。你现在不需要任何 for 循环。试试看，让我知道它是否有效！

此外，其中包含 3 个数字的 elif 语句的编写方式可能不起作用。我为你重写了它们：

elif (second_half > 0) and (second_half < 30) and (second_half != 15):
    second_half_2 = f"{second_half} minutes past "
elif (second_half > 30) and (second_half < 59) and (second_half != 45):
    second_half_2 = f"{60 - second_half} minutes to "