0-23小时军用时钟到标准时间(hh:mm)

Question

我该如何处理这个过程？假设有人输入"3:5",我应该能够得到输出"3:05 AM".或者,如果有人输入"00:00",我应该得到"12:00 AM".如果时间格式只是输入为"hhmm",我会意识到过多的方法,但输入的范围可以是"h:m","hh:m","hh:m"和"hh:毫米".所以它非常具有情境性.

def timeConvert():
    miliTime = int(input("Enter a time in hh:mm (military) format: "))
    miliTime.split(":")
    if len(miliTime) == 3:
        hours = miliTime[0]
        minutes = miliTime[2]
        if hours < 0:
            print("Hours can't be less than 0.")
    elif len(miliTime) == 4:
        if miliTime[0:2] >= 10:
            hours = militime[0:2]
            minutes = militime[3]
            if minutes < 0:
                print("Minutes can't be less than 0.")
            if minutes < 10:
                minutes = 0 + minutes
        else:
            hours = miliTime[0]
            minutes = militime[2:]
            if minutes >= 60:
                print("Too big of a number for minutes.")
    else:
        hours = miliTime[0:2]
        minutes = miliTime[3:]
    setting = AM
    if hours > 12:
        setting = PM
        hours -= 12
    print(hours + ":" + minutes + setting)

timeConvert()

Answer 1

使用datetime模块strptime()将字符串作为日期时间对象读取,然后将其转换回字符串strftime(),例如 -

>>> datetime.datetime.strptime('3:5','%H:%M').strftime('%I:%M %p')
'03:05 AM'
>>> datetime.datetime.strptime('18:30','%H:%M').strftime('%I:%M %p')
'06:30 PM'
>>> datetime.datetime.strptime('00:00','%H:%M').strftime('%I:%M %p')
'12:00 AM'

格式说明 -

%H - 小时(24小时fomrat)

%M - 分钟

%I - 小时(12小时格式)

%p - AM或PM

功能看起来像 -

def timeConvert():
    import datetime
    ds = input("Enter a time in hh:mm (military) format: ")
    newds = datetime.datetime.strptime(ds, ,'%H:%M').strftime('%I:%M %p')
    print(newds)

Answer 2

按 分割输入:，并最后使用格式化输出：

def timeConvert():
    miliTime = input("Enter a time in hh:mm (military) format: ")
    hours, minutes = miliTime.split(":")
    hours, minutes = int(hours), int(minutes)
    setting = "AM"
    if hours > 12:
        setting = "PM"
        hours -= 12
    print(("%02d:%02d" + setting) % (hours, minutes))