将UTC日期转换为毫秒 [英] Convert UTC date into milliseconds

问题描述

我对当前的UTC时间毫秒不感兴趣，也不需要弄乱时区。我的原始日期已经存储为UTC时间戳。

I am not interested in what the current UTC time is in milliseconds, nor do I need to mess with timezones. My original date is already stored as a UTC timestamp.

我在UTC时间存储在数据库中的日期，2012-06-14 05:01:25 。
我对日期时间不感兴趣，只是它的日期部分。因此，在使用Java检索日期并排除小时，分钟和秒后 - 我留下了2012-06-14。

I have a date stored in a database in UTC time, "2012-06-14 05:01:25". I am not interested in the datetime, but just the date portion of the it. So, after retrieving the date in Java, and excluding the hours, minutes, and seconds - I am left with "2012-06-14".

我如何转换这是UTC毫秒？

How can I convert this into UTC milliseconds?

推荐答案

编辑：我错过了忽略一天中的时间部分。它现在存在，但接近结束......

I'd missed the "ignoring the time of day" part. It's now present, but near the end...

最简单的方法可能是使用 SimpleDateFormat ，已适当设置时区：

The simplest approach is probably to use SimpleDateFormat, having set the time zone appropriately:

SimpleDateFormat format = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss", Locale.US);
format.setTimeZone(TimeZone.getTimeZone("UTC"));

Date date = format.parse(text);
long millis = date.getTime();

（设置时区是这里的重要位，否则它将解释为的值本地时区。）

(Setting the time zone is the important bit here, as otherwise it will interpret the value to be in the local time zone.)

或者，如果您正在做的事情比这更简单，请使用Joda Time 这是一个更好的日期/时间API。特别是， SimpleDateFormat 不是线程安全的，而 DateTimeFormatter 是：

Alternatively, if you're doing anything less trivial than this, use Joda Time which is a much better date/time API. In particular, SimpleDateFormat isn't thread-safe whereas DateTimeFormatter is:

// This can be reused freely across threads after construction.
DateTimeFormatter formatter = DateTimeFormat.forPattern("yyyy-MM-dd HH:mm:ss")
    .withLocale(Locale.US)
    .withZoneUTC();

// Option 1
DateTime datetime = formatter.parseDateTime(text);
long millis = dateTime.getMillis();

// Option 2, more direct, but harder to diagnose errors
long millis = formatter.parseMillis(text);

现在到目前为止，我们已经解析了整个caboodle。忽略日期部分的最简单方法就是将其四舍五入 - 毕竟，Java没有观察到闰秒，所以我们可以截断它：

Now so far, we've parsed the whole whole caboodle. The easiest way of ignoring the date part is just to round it off - after all, Java doesn't observe leap seconds, so we can just truncate it:

long millisPerDay = 24L * 60L * 60L * 1000L; // Or use TimeUnit
long dayMillis = (millis / millisPerDay) * millisPerDay;

这将向1970年前进，所以如果你有之前的日期 1970年它将转到当天的结束 - 但我怀疑这不太可能是一个问题。

That will "round towards 1970" so if you have a date before 1970 it will round to the end of the day - but I suspect that's unlikely to be a problem.

使用Joda Time版本你可以只需使用它：

With the Joda Time version you could just use this instead:

DateTime dateTime = formatter.parseDateTime(text);
long millis = dateTime.toLocalDate().getLocalMillis();

我个人不只想带一个子串。即使你实际上对保留小时/分钟/秒感兴趣，我认为解析你已经给出的内容并且然后丢弃信息是合适的。除此之外，它会使您的代码在错误数据下失败，例如

I would personally not go with the idea of just taking a substring. Even though you're not actually interested in preserving the hour/minute/second, I think it's appropriate to parse what you've been given and then throw away information. Aside from anything else, it makes your code fail appropriately with bad data, e.g.

"2012-06-100"

或

"2012-06-14 25:01:25"

表示无论是什么为您提供数据，并且发现它是好的，而不是盲目地继续，只因为前10个字符是可以的。

indicate problems in whatever's supplying you data, and it's good to spot that rather than to continue blindly just because the first 10 characters are okay.

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