将纪元以来的天数转换为自纪元以来的秒数

At my new workplace, they represent a lot of dates as "days since epoch" (which I will hereafter call DSE). I'm running into issues in JavaScript converting from DSE to seconds since epoch (UNIX timestamps). Here's my function to do the conversion:

在我的新工作场所,他们代表了许多日期为“自纪元以来的日子”(我将在此后称之为DSE)。自从epoch(UNIX时间戳)以来,我遇到了从DSE转换为秒的问题。这是我进行转换的功能:

function daysToTimestamp(days) {
    return Math.round(+days * 86400);
}

By way of example, when I pass in 13878 (expecting that this represents January 1, 2008), I get back 1199059200, not 1199098800 as I expect. Why?

举例来说,当我通过13878(期望这代表2008年1月1日)时,我得到了1199059200,而不是我预期的1199098800。为什么?

5 个解决方案

#1

1199059200 represents December 31 2007 in UTC. Sample Python session:

1199059200表示2007年12月31日UTC。示例Python会话:

>>> import time
>>> time.gmtime(1199059200)
(2007, 12, 31, 0, 0, 0, 0, 365, 0)

Remember that all time_t values are against UTC. :-) You have to adjust accordingly to your timezone.

请记住,所有time_t值都是针对UTC的。 :-)你必须根据你的时区进行调整。

Edit: Since you and I are both in New Zealand, here's how you might have got the 1199098800 value:

编辑:既然你和我都在新西兰,你可能会得到1199098800的价值:

>>> time.localtime(1199098800)
(2008, 1, 1, 0, 0, 0, 1, 1, 1)

This is so because in New Year (summer in New Zealand), the timezone here is +1300. Do the maths and see. :-)

这是因为在新年(新西兰的夏天),这里的时区是+1300。做数学并看看。 :-)

For January 1 2008 in UTC, add 86400 to 1199059200, and get 1199145600.

对于2008年1月1日的UTC,添加86400到1199059200,得到1199145600。

>>> time.gmtime(1199145600)
(2008, 1, 1, 0, 0, 0, 1, 1, 0)

#2

It is because it is neither a linear representation of time nor a true representation of UTC (though it is frequently mistaken for both) as the times it represents are UTC but it has no way of representing UTC leap seconds

这是因为它既不是时间的线性表示也不是UTC的真实表示(虽然它经常被误认为是两者),因为它表示的时间是UTC,但它无法表示UTC闰秒

http://en.wikipedia.org/wiki/Unix_time

#3

Unix times (time_t) are represented in seconds since Jan 1, 1970 not milliseconds.

Unix时间(time_t)以1970年1月1日以秒为单位表示,而不是毫秒。

I imagine what you are seeing is a difference in timezone. The delta you have is 11 hours, how are you getting the expected value?

我想你所看到的是时区的差异。您拥有的delta是11个小时,您是如何获得预期值的?

#4

Because 1199098800/86400 = 13878.4583333333333 (with the 3 repeating forever), not 13878.0. It gets rounded to 13878.0 since it's being stored as an integer. If you want to see the difference it makes, try this: .4583333333333*86400 = 39599.99999999712. Even that makes it slightly incorrect, but this is where the discrepancy comes from, as 1199098800-1199059200=35600.

因为1199098800/86400 = 13878.4583333333333(与3重复永远),而不是13878.0。它被舍入到13878.0,因为它被存储为整数。如果你想看到它的差异,试试这个:.4583333333333 * 86400 = 39599.99999999712。即使这样也会略微不正确,但这就是差异所在,如1199098800-1199059200 = 35600。

#5

-1

You should multiply by 86400000

你应该乘以86400000

1 day = 24 hours * 60 minutes * 60 seconds * 1000 milliseconds = 86400000

1天= 24小时* 60分钟* 60秒* 1000毫秒= 86400000

#1