将unix时间戳转换为YYYY-MM-DD HH:MM:SS

Question

我有一个Unix时间戳,我需要从中获取单独的年,月,日,小时,分钟和秒.我在数学课上从来都不是很好,所以我想知道你们是否可以帮我一点:)

我必须自己做所有事情(没有time.h函数).语言是C.

Answer 1

免责声明:以下代码不考虑闰年 或闰秒 [Unix时间不考虑闰秒.无论如何,他们被高估了.-Ed].另外,我没有测试它,所以可能有bug.它可能会踢你的猫并侮辱你的母亲.祝你今天愉快.

让我们尝试一下psuedocode(Python,真的):

# Define some constants here...

# You'll have to figure these out.  Don't forget about February (leap years)...
secondsPerMonth = [ SECONDS_IN_JANUARY, SECONDS_IN_FEBRUARY, ... ]

def formatTime(secondsSinceEpoch):
    # / is integer division in this case.
    # Account for leap years when you get around to it :)
    year = 1970 + secondsSinceEpoch / SECONDS_IN_YEAR
    acc = secondsSinceEpoch - year * SECONDS_IN_YEAR

    for month in range(12):
        if secondsPerMonth[month] < acc:
            acc -= month
            month += 1

    month += 1

    # Again, / is integer division.
    days = acc / SECONDS_PER_DAY
    acc -= days * SECONDS_PER_DAY

    hours = acc / SECONDS_PER_HOUR
    acc -= hours * SECONDS_PER_HOUR

    minutes = acc / SECONDS_PER_MINUTE
    acc -= minutes * SECONDS_PER_MINUTE

    seconds = acc

    return "%d-%d-%d %d:%d%d" % (year, month, day, hours, minutes, seconds)

如果我搞砸了,请告诉我.在C中这样做不应该太难.