Java 8:以多个单位表示的两个LocalDateTime之间的差异 [英] Java 8: Difference between two LocalDateTime in multiple units

问题描述

我正在尝试计算两个LocalDateTime之间的差异.

I am trying to calculate the difference between two LocalDateTime.

输出的格式必须为y years m months d days h hours m minutes s seconds.这是我写的:

The output needs to be of the format y years m months d days h hours m minutes s seconds. Here is what I have written:

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.我已经从该网站(值为12/16/1984 07:45:55和09/09/2014 19:46:45)检查了我的结果.以下屏幕截图显示了输出:

The output that I am getting is 29 years 8 months 24 days 12 hours 0 minutes 50 seconds. I have checked my result from this website (with values 12/16/1984 07:45:55 and 09/09/2014 19:46:45). The following screenshot shows the output:

我很确定月份值后面的字段在我的代码中出了错.任何建议都会非常有帮助.

I am pretty sure that the fields after the month value is coming wrong from my code. Any suggestion would be very helpful.

我已经从另一个网站测试了我的结果，但是得到的结果却有所不同.在这里是:

I have tested my result from another website and the result I got is different. Here it is: Calculate duration between two dates (result: 29 years, 8 months, 24 days, 12 hours, 0 minutes and 50 seconds).

由于我从两个不同的站点获得了两个不同的结果，所以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:

Since I got two different results from two different sites, I am wondering if the algorithm of my calculation is legitimate or not. If I use following two LocalDateTime objects:

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

然后输出即将到来:29 years 8 months 25 days -1 hours -5 minutes -10 seconds.

从此

From this link it should be 29 years 8 months 24 days 22 hours, 54 minutes and 50 seconds. So the algorithm needs to handle the negative numbers too.

请注意，问题不在于哪个网站给我带来了什么结果，我需要了解正确的算法并需要获得正确的结果.

推荐答案

不幸的是，似乎也没有跨时间的周期类，因此您可能必须自己进行计算.

Unfortunately, there doesn't seem to be a period class that spans time as well, so you might have to do the calculations on your own.

幸运的是，日期和时间类具有许多实用程序方法，可以在某种程度上简化该过程.这是一种计算差异的方法，尽管不一定最快:

Fortunately, the date and time classes have a lot of utility methods that simplify that to some degree. Here's a way to calculate the difference although not necessarily the fastest:

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.

基本思想是:创建一个临时的开始日期并获得整年的结束时间.然后，根据年数调整该日期，以使开始日期少于结束日期.对每个时间单位都按降序重复该操作.

The basic idea is this: create a temporary start date and get the full years to the end. Then adjust that date by the number of years so that the start date is less then a year from the end. Repeat that for each time unit in descending order.

最后一个免责声明:我没有考虑不同的时区(两个日期都应该在同一时区中)，我也没有测试/检查夏令时或其他更改日历(例如萨摩亚中的时区更改)会影响此计算.因此，请谨慎使用.

Finally a disclaimer: I didn't take different timezones into account (both dates should be in the same timezone) and I also didn't test/check how daylight saving time or other changes in a calendar (like the timezone changes in Samoa) affect this calculation. So use with care.

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