转换unix时间戳，以YYYY-MM-DD HH：MM：SS [英] Converting unix timestamp to YYYY-MM-DD HH:MM:SS

问题描述

我有一个Unix时间戳，我需要得到个人的年，月，日，时，分，秒值。所以我在想，如果你们可以帮助我一点我从来没有在数学课非常好：）

I have a Unix timestamp that I need to get the individual year, month, day, hour, minute and second values from. I never was very good in math class so I was wondering if you guys could help me out a little :)

我做的一切我自己（没有time.h中的功能）。语言是C。

I have to do everything myself (no time.h functions). The language is C.

推荐答案

免责声明：以下code不占的闰年 <击>或闰秒 的 [Unix时间呢不占闰秒。他们是被高估的，反正。 -ed] 的。另外，我没有测试，所以可能有错误。它可能会踢你的猫和侮辱你的母亲。有一个愉快的一天。

Disclaimer: The following code does not account for leap years or leap seconds [Unix time does not account for leap seconds. They're overrated, anyway. -Ed]. Also, I did not test it, so there may be bugs. It may kick your cat and insult your mother. Have a nice day.

让我们尝试一点点伪code（Python的，真的）：

Let's try a little psuedocode (Python, really):

# 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)

如果我疯玩了，请让我知道。用C这样做应该不会太困难得多。

If I goofed up, please let me know. Doing this in C shouldn't be too much harder.

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