我相信下面的方法給出了你的Joda時(shí)間解的等價(jià)物。

private static final LocalTime START = LocalTime.of(18, 0);

private static final LocalTime END = LocalTime.of(8, 0);

public static Duration overlap(ZonedDateTime currentStart, ZonedDateTime currentEnd) {

ZonedDateTime singleIntervalStart = currentStart.with(START);

ZonedDateTime singleIntervalEnd = currentStart.plusDays(1).with(END);

if (currentEnd.isBefore(singleIntervalStart)) {

// no overlap

return Duration.ZERO;

}

ZonedDateTime overlapStart = currentStart.isBefore(singleIntervalStart)

? singleIntervalStart : currentStart;

ZonedDateTime overlapEnd = currentEnd.isBefore(singleIntervalEnd)

? currentEnd : singleIntervalEnd;

return Duration.between(overlapStart, overlapEnd);

}

為了使用您問(wèn)題中的時(shí)間戳進(jìn)行嘗試,我使用以下實(shí)用程序方法:

private static void demo(String from, String to) {

ZoneId zone = ZoneId.of("Atlantic/Stanley");

Duration overlapDuration = overlap(LocalDateTime.parse(from).atZone(zone),

LocalDateTime.parse(to).atZone(zone));

System.out.println("" + from + " - " + to + ": " + overlapDuration);

}

現(xiàn)在我這樣稱呼它:

demo("2018-01-02T14:59:18.922", "2018-01-02T14:59:38.804");

demo("2018-01-02T18:32:59.348", "2018-01-02T20:30:41.192");

demo("2018-01-02T01:54:59.363", "2018-01-02T01:54:59.363");

demo("2018-01-03T00:10:38.831", "2018-01-03T00:11:53.103");

輸出為:

2018-01-02T14:59:18.922 - 2018-01-02T14:59:38.804: PT0S

2018-01-02T18:32:59.348 - 2018-01-02T20:30:41.192: PT1H57M41.844S

2018-01-02T01:54:59.363 - 2018-01-02T01:54:59.363: PT0S

2018-01-03T00:10:38.831 - 2018-01-03T00:11:53.103: PT0S

在第一個(gè)示例中,14:59在18:00之前,因此結(jié)果是0的重疊。在第二個(gè)示例中,整個(gè)間隔被計(jì)算為重疊(近2小時(shí))。請(qǐng)注意,在最后兩個(gè)示例中,沒(méi)有報(bào)告重疊,因?yàn)闀r(shí)間在18:00之前很多小時(shí)。我不確定這是否是你想要的,因?yàn)樘┪钍繄?bào)也在08:00之前。

總結(jié)

以上是生活随笔為你收集整理的java 计算两个时间戳_Java时间戳计算重叠持续时间与间隔的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問(wèn)題。

如果覺(jué)得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。