Joda-Time

Question

我正在尝试计算两个LocalDateTime之间的差异。

输出必须是y years m months d days h hours m minutes s seconds格式。这是我写的：

import java.time.Duration;
import java.time.Instant;
import java.time.LocalDateTime;
import java.time.Period;
import java.time.ZoneId;

public class Main {

    static final int MINUTES_PER_HOUR = 60;
    static final int SECONDS_PER_MINUTE = 60;
    static final int SECONDS_PER_HOUR = SECONDS_PER_MINUTE * MINUTES_PER_HOUR;

    public static void main(String[] args) {
        LocalDateTime toDateTime = LocalDateTime.of(2014, 9, 9, 19, 46, 45);
        LocalDateTime fromDateTime = LocalDateTime.of(1984, 12, 16, 7, 45, 55);

        Period period = getPeriod(fromDateTime, toDateTime);
        long time[] = getTime(fromDateTime, toDateTime);

        System.out.println(period.getYears() + " years " + 
                period.getMonths() + " months " + 
                period.getDays() + " days " +
                time[0] + " hours " +
                time[1] + " minutes " +
                time[2] + " seconds.");


    }

    private static Period getPeriod(LocalDateTime dob, LocalDateTime now) {
        return Period.between(dob.toLocalDate(), now.toLocalDate());
    }

    private static long[] getTime(LocalDateTime dob, LocalDateTime now) {
        LocalDateTime today = LocalDateTime.of(now.getYear(),
                now.getMonthValue(), now.getDayOfMonth(), dob.getHour(), dob.getMinute(), dob.getSecond());
        Duration duration = Duration.between(today, now);

        long seconds = duration.getSeconds();

        long hours = seconds / SECONDS_PER_HOUR;
        long minutes = ((seconds % SECONDS_PER_HOUR) / SECONDS_PER_MINUTE);
        long secs = (seconds % SECONDS_PER_MINUTE);

        return new long[]{hours, minutes, secs};
    }
}

我得到的输出是29 years 8 months 24 days 12 hours 0 minutes 50 seconds。我已从此website检查了我的结果（值为12/16/1984 07:45:55和09/09/2014 19:46:45）。以下屏幕截图显示了输出：

Epoch Converter

我很确定月份值之后的字段从我的代码中出错了。任何建议都会非常有用。

更新

我已经从另一个网站测试了我的结果，我得到的结果是不同的。这是：Calculate duration between two dates（结果：29年，8个月，24天，12小时，0分50秒）。

更新

由于我从两个不同的网站得到两个不同的结果，我想知道我的计算算法是否合法。如果我使用以下两个LocalDateTime对象：

LocalDateTime toDateTime = LocalDateTime.of(2014, 9, 10, 6, 40, 45);
LocalDateTime fromDateTime = LocalDateTime.of(1984, 12, 16, 7, 45, 55);

然后输出结束：29 years 8 months 25 days -1 hours -5 minutes -10 seconds.

从这link开始，它应该是29 years 8 months 24 days 22 hours, 54 minutes and 50 seconds。所以算法也需要处理负数。

请注意，问题不在于哪个网站给了我什么结果，我需要知道正确的算法并且需要有正确的结果。

Answer 1

我发现最好的方法是使用 ChronoUnit。

long minutes = ChronoUnit.MINUTES.between(fromDate, toDate);
long hours = ChronoUnit.HOURS.between(fromDate, toDate);

其他文档在此处：https://docs.oracle.com/javase/tutorial/datetime/iso/period.html

Answer 2

不幸的是，似乎并不是一个跨越时间的句号类，所以你可能需要自己进行计算。

幸运的是，日期和时间类有很多实用方法可以在某种程度上简化它。这是计算差异的一种方法，但不一定是最快的：

LocalDateTime fromDateTime = LocalDateTime.of(1984, 12, 16, 7, 45, 55);
LocalDateTime toDateTime = LocalDateTime.of(2014, 9, 10, 6, 40, 45);

LocalDateTime tempDateTime = LocalDateTime.from( fromDateTime );

long years = tempDateTime.until( toDateTime, ChronoUnit.YEARS);
tempDateTime = tempDateTime.plusYears( years );

long months = tempDateTime.until( toDateTime, ChronoUnit.MONTHS);
tempDateTime = tempDateTime.plusMonths( months );

long days = tempDateTime.until( toDateTime, ChronoUnit.DAYS);
tempDateTime = tempDateTime.plusDays( days );


long hours = tempDateTime.until( toDateTime, ChronoUnit.HOURS);
tempDateTime = tempDateTime.plusHours( hours );

long minutes = tempDateTime.until( toDateTime, ChronoUnit.MINUTES);
tempDateTime = tempDateTime.plusMinutes( minutes );

long seconds = tempDateTime.until( toDateTime, ChronoUnit.SECONDS);

System.out.println( years + " years " + 
        months + " months " + 
        days + " days " +
        hours + " hours " +
        minutes + " minutes " +
        seconds + " seconds.");

//prints: 29 years 8 months 24 days 22 hours 54 minutes 50 seconds.

基本思路是这样的：创建一个临时的开始日期，并将整年带到最后。然后按年数调整该日期，以便开始日期少于结束一年。对每个时间单位按降序重复该操作。

最后一个免责声明：我没有考虑不同的时区（两个日期应该在同一时区），我也没有测试/检查日历中的夏令时或其他变化（比如萨摩亚的时区变化）会影响这个计算。所以要小心使用。

Answer 3

这里有一个使用Duration和TimeUnit获取'hh：mm：ss'格式的例子。

Duration dur = Duration.between(localDateTimeIni, localDateTimeEnd);
long millis = dur.toMillis();

String.format("%02d:%02d:%02d", 
        TimeUnit.MILLISECONDS.toHours(millis),
        TimeUnit.MILLISECONDS.toMinutes(millis) - 
        TimeUnit.HOURS.toMinutes(TimeUnit.MILLISECONDS.toHours(millis)),
        TimeUnit.MILLISECONDS.toSeconds(millis) - 
        TimeUnit.MINUTES.toSeconds(TimeUnit.MILLISECONDS.toMinutes(millis)));

Answer 4

应该更简单！

Duration.between(startLocalDateTime, endLocalDateTime).toMillis();

Answer 5

Groovy中的@Thomas版本采用列表中所需的单位而不是硬编码值。这个实现（可以很容易地移植到Java - 我使函数声明显式化）使得Thomas更可重用。

def fromDateTime = LocalDateTime.of(1968, 6, 14, 0, 13, 0)
def toDateTime = LocalDateTime.now()
def listOfUnits = [
    ChronoUnit.YEARS, ChronoUnit.MONTHS, ChronoUnit.DAYS,
    ChronoUnit.HOURS, ChronoUnit.MINUTES, ChronoUnit.SECONDS,
    ChronoUnit.MILLIS]

println calcDurationInTextualForm(listOfUnits, fromDateTime, toDateTime)    

String calcDurationInTextualForm(List<ChronoUnit> listOfUnits, LocalDateTime ts, LocalDateTime to)
{
    def result = []

    listOfUnits.each { chronoUnit ->
        long amount = ts.until(to, chronoUnit)
        ts = ts.plus(amount, chronoUnit)

        if (amount) {
            result << "$amount ${chronoUnit.toString()}"
        }
    }

    result.join(', ')
}

在撰写本文时，上面的代码返回47 Years, 8 Months, 9 Days, 22 Hours, 52 Minutes, 7 Seconds, 140 Millis。并且，对于@Gennady Kolomoets输入，代码返回23 Hours。

当您提供单位列表时，必须按单位大小排序（最大的首先）：

def listOfUnits = [ChronoUnit.WEEKS, ChronoUnit.DAYS, ChronoUnit.HOURS]
// returns 2495 Weeks, 3 Days, 8 Hours

Answer 6

这是对您的问题的一个非常简单的答案。它有效。

import java.time.*;
import java.util.*;
import java.time.format.DateTimeFormatter;
public class MyClass {
    public static void main(String args[]) {
       DateTimeFormatter T = DateTimeFormatter.ofPattern("dd/MM/yyyy HH:mm");
       Scanner h = new Scanner(System.in);

       System.out.print("Enter date of birth[dd/mm/yyyy hh:mm]: ");
       String b = h.nextLine();

       LocalDateTime bd = LocalDateTime.parse(b,T);
       LocalDateTime cd = LocalDateTime.now();

       int hr = cd.getHour() - bd.getHour();
       int mn = cd.getMinute() - bd.getMinute();

       Period time = Period.between(bd.toLocalDate(),cd.toLocalDate());

       System.out.print("Age is: "+time.getYears()+ " years,"+time.getMonths()+ " months, " +time.getDays()+ " days, "+hr+ " hours, " +mn+ " minutes old");
    }
}

Answer 7

Tapas Bose代码和Thomas代码存在一些问题。如果时间差异为负，则数组获得负值。例如，如果

LocalDateTime toDateTime = LocalDateTime.of(2014, 9, 10, 6, 46, 45);
LocalDateTime fromDateTime = LocalDateTime.of(2014, 9, 9, 7, 46, 45);

它返回0年0个月1天-1小时0分0秒。

我认为正确的输出是：0年0个月0天23小时0分0秒。

我建议在LocalDate和LocalTime实例上分离LocalDateTime实例。之后，我们可以获得Java 8 Period和Duration实例。持续时间实例按天数和全天时间值（<24h）分开，随后校正周期值。当第二个LocalTime值在firstLocalTime值之前时，有必要将句点缩短一天。

这是我计算LocalDateTime差异的方法：

private void getChronoUnitForSecondAfterFirst(LocalDateTime firstLocalDateTime, LocalDateTime secondLocalDateTime, long[] chronoUnits) {
    /*Separate LocaldateTime on LocalDate and LocalTime*/
    LocalDate firstLocalDate = firstLocalDateTime.toLocalDate();
    LocalTime firstLocalTime = firstLocalDateTime.toLocalTime();

    LocalDate secondLocalDate = secondLocalDateTime.toLocalDate();
    LocalTime secondLocalTime = secondLocalDateTime.toLocalTime();

    /*Calculate the time difference*/
    Duration duration = Duration.between(firstLocalDateTime, secondLocalDateTime);
    long durationDays = duration.toDays();
    Duration throughoutTheDayDuration = duration.minusDays(durationDays);
    Logger.getLogger(PeriodDuration.class.getName()).log(Level.INFO,
            "Duration is: " + duration + " this is " + durationDays
            + " days and " + throughoutTheDayDuration + " time.");

    Period period = Period.between(firstLocalDate, secondLocalDate);

    /*Correct the date difference*/
    if (secondLocalTime.isBefore(firstLocalTime)) {
        period = period.minusDays(1);
        Logger.getLogger(PeriodDuration.class.getName()).log(Level.INFO,
                "minus 1 day");
    }

    Logger.getLogger(PeriodDuration.class.getName()).log(Level.INFO,
            "Period between " + firstLocalDateTime + " and "
            + secondLocalDateTime + " is: " + period + " and duration is: "
            + throughoutTheDayDuration
            + "\n-----------------------------------------------------------------");

    /*Calculate chrono unit values and  write it in array*/
    chronoUnits[0] = period.getYears();
    chronoUnits[1] = period.getMonths();
    chronoUnits[2] = period.getDays();
    chronoUnits[3] = throughoutTheDayDuration.toHours();
    chronoUnits[4] = throughoutTheDayDuration.toMinutes() % 60;
    chronoUnits[5] = throughoutTheDayDuration.getSeconds() % 60;
}

上述方法可用于计算任何本地日期和时间值的差异，例如：

public long[] getChronoUnits(String firstLocalDateTimeString, String secondLocalDateTimeString) {
    DateTimeFormatter formatter = DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss");

    LocalDateTime firstLocalDateTime = LocalDateTime.parse(firstLocalDateTimeString, formatter);
    LocalDateTime secondLocalDateTime = LocalDateTime.parse(secondLocalDateTimeString, formatter);

    long[] chronoUnits = new long[6];
    if (secondLocalDateTime.isAfter(firstLocalDateTime)) {
        getChronoUnitForSecondAfterFirst(firstLocalDateTime, secondLocalDateTime, chronoUnits);
    } else {
        getChronoUnitForSecondAfterFirst(secondLocalDateTime, firstLocalDateTime, chronoUnits);
    }
    return chronoUnits;
}

为上述方法编写单元测试很方便（它们都是PeriodDuration类成员）。这是代码：

@RunWith(Parameterized.class)
public class PeriodDurationTest {

private final String firstLocalDateTimeString;
private final String secondLocalDateTimeString;
private final long[] chronoUnits;

public PeriodDurationTest(String firstLocalDateTimeString, String secondLocalDateTimeString, long[] chronoUnits) {
    this.firstLocalDateTimeString = firstLocalDateTimeString;
    this.secondLocalDateTimeString = secondLocalDateTimeString;
    this.chronoUnits = chronoUnits;
}

@Parameters
public static Collection<Object[]> periodValues() {
    long[] chronoUnits0 = {0, 0, 0, 0, 0, 0};
    long[] chronoUnits1 = {0, 0, 0, 1, 0, 0};
    long[] chronoUnits2 = {0, 0, 0, 23, 0, 0};
    long[] chronoUnits3 = {0, 0, 0, 1, 0, 0};
    long[] chronoUnits4 = {0, 0, 0, 23, 0, 0};
    long[] chronoUnits5 = {0, 0, 1, 23, 0, 0};
    long[] chronoUnits6 = {29, 8, 24, 12, 0, 50};
    long[] chronoUnits7 = {29, 8, 24, 12, 0, 50};
    return Arrays.asList(new Object[][]{
        {"2015-09-09 21:46:44", "2015-09-09 21:46:44", chronoUnits0},
        {"2015-09-09 21:46:44", "2015-09-09 22:46:44", chronoUnits1},
        {"2015-09-09 21:46:44", "2015-09-10 20:46:44", chronoUnits2},
        {"2015-09-09 21:46:44", "2015-09-09 20:46:44", chronoUnits3},
        {"2015-09-10 20:46:44", "2015-09-09 21:46:44", chronoUnits4},
        {"2015-09-11 20:46:44", "2015-09-09 21:46:44", chronoUnits5},
        {"1984-12-16 07:45:55", "2014-09-09 19:46:45", chronoUnits6},
        {"2014-09-09 19:46:45", "1984-12-16 07:45:55", chronoUnits6}
    });
}

@Test
public void testGetChronoUnits() {
    PeriodDuration instance = new PeriodDuration();
    long[] expResult = this.chronoUnits;
    long[] result = instance.getChronoUnits(this.firstLocalDateTimeString, this.secondLocalDateTimeString);
    assertArrayEquals(expResult, result);
}

}

无论第一个LocalDateTime的值是否在任何LocalTime值之前，所有测试都是成功的。

Answer 8

五年后，我回答了我的问题。我认为持续时间为负数的问题可以通过简单的修正即可解决：

LocalDateTime fromDateTime = LocalDateTime.of(2014, 9, 9, 7, 46, 45);
LocalDateTime toDateTime = LocalDateTime.of(2014, 9, 10, 6, 46, 45);

Period period = Period.between(fromDateTime.toLocalDate(), toDateTime.toLocalDate());
Duration duration = Duration.between(fromDateTime.toLocalTime(), toDateTime.toLocalTime());

if (duration.isNegative()) {
    period = period.minusDays(1);
    duration = duration.plusDays(1);
}
long seconds = duration.getSeconds();
long hours = seconds / SECONDS_PER_HOUR;
long minutes = ((seconds % SECONDS_PER_HOUR) / SECONDS_PER_MINUTE);
long secs = (seconds % SECONDS_PER_MINUTE);
long time[] = {hours, minutes, secs};
System.out.println(period.getYears() + " years "
            + period.getMonths() + " months "
            + period.getDays() + " days "
            + time[0] + " hours "
            + time[1] + " minutes "
            + time[2] + " seconds.");

注意：站点https://www.epochconverter.com/date-difference现在可以正确计算时差。

谢谢大家的讨论和建议。

Answer 9

Joda-Time

由于许多答案都需要 API 26 支持，而我的min API是23，因此我通过以下代码解决了该问题：

import org.joda.time.Days

LocalDate startDate = Something
LocalDate endDate = Something
// The difference would be exclusive of both dates, 
// so in most of use cases we may need to increment it by 1
Days.daysBetween(startDate, endDate).days

Answer 10

TL; DR

Duration duration = Duration.between(start, end);
duration = duration.minusDays(duration.toDaysPart()); // essentially "duration (mod 1 day)"
Period period = Period.between(start.toLocalDate(), end.toLocalDate());

，然后使用方法period.getYears()，period.getMonths()，period.getDays()，duration.toHoursPart()，duration.toMinutesPart()，duration.toSecondsPart()。

扩展答案

我将回答原始问题，即如何获取两个LocalDateTimes之间的年，月，日，小时和分钟之间的时差，以使所有不同单位的值等于总时间差，因此每个单位的值小于下一个较大的单位，即minutes < 60，hours < 24等。

给出两个LocalDateTimes start和end，例如

LocalDateTime start = LocalDateTime.of(2019, 11, 29, 17, 15);
LocalDateTime end = LocalDateTime.of(2020, 11, 30, 18, 44);

我们可以用Duration表示两者之间的绝对时间跨度-也许使用Duration.between(start, end)。但是，我们可以从Duration中提取的最大单位是天（相当于24h的时间单位），请参阅下面的注释以获取解释。要使用较大的单位（月，年），我们可以用一对{Duration，Period）来表示此Duration，其中Period可以精确地测量差异天，而Duration代表剩余时间：

Duration duration = Duration.between(start, end);
duration = duration.minusDays(duration.toDaysPart()); // essentially "duration (mod 1 day)"
Period period = Period.between(start.toLocalDate(), end.toLocalDate());

现在，我们可以简单地使用Period和Duration上定义的方法来提取各个单位：

System.out.printf("%d years, %d months, %d days, %d hours, %d minutes, %d seconds",
        period.getYears(), period.getMonths(), period.getDays(), duration.toHoursPart(),
        duration.toMinutesPart(), duration.toSecondsPart());

1 years, 0 months, 1 days, 1 hours, 29 minutes, 0 seconds

或使用默认格式：

System.out.println(period + " + " + duration);

P1Y1D + PT1H29M

注意年月日

请注意，在java.time的概念中，“月”或“年”之类的“单位”并不代表固定的绝对时间值，它们与日期和日历有关，因为以下示例说明了这一点：

LocalDateTime
        start1 = LocalDateTime.of(2020, 1, 1, 0, 0),
        end1 = LocalDateTime.of(2021, 1, 1, 0, 0),
        start2 = LocalDateTime.of(2021, 1, 1, 0, 0),
        end2 = LocalDateTime.of(2022, 1, 1, 0, 0);
System.out.println(Period.between(start1.toLocalDate(), end1.toLocalDate()));
System.out.println(Duration.between(start1, end1).toDays());
System.out.println(Period.between(start2.toLocalDate(), end2.toLocalDate()));
System.out.println(Duration.between(start2, end2).toDays());

P1Y
366
P1Y
365

Java 8：计算两个LocalDateTime之间的差异

更新

更新

10 个答案:

Joda-Time

TL; DR

扩展答案

注意年月日