Java 8:多个单位中两个 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.

输出的格式需要为y 年 m 月 d 天 h 小时 m 分 s 秒.这是我写的:

import java.time.Duration;导入 java.time.Instant;导入 java.time.LocalDateTime;导入 java.time.Period;导入 java.time.ZoneId;公共课主要{静态最终 int MINUTES_PER_HOUR = 60;静态最终 int SECONDS_PER_MINUTE = 60;静态最终 int SECONDS_PER_HOUR = SECONDS_PER_MINUTE * MINUTES_PER_HOUR;公共静态无效主(字符串 [] 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() + "年" +period.getMonths() + 月" +period.getDays() + " 天 " +时间[0] + "小时" +时间[1] + "分钟" +时间[2] + "秒");}private static Period getPeriod(LocalDateTime dob, LocalDateTime now) {return Period.between(dob.toLocalDate(), now.toLocalDate());}private static long[] getTime(LocalDateTime dob, LocalDateTime now) {今天的 LocalDateTime = LocalDateTime.of(now.getYear(),now.getMonthValue(), now.getDayOfMonth(), dob.getHour(), dob.getMinute(), dob.getSecond());Duration 持续时间 = Duration.between(today, now);长秒 = duration.getSeconds();长时间 = 秒/SECONDS_PER_HOUR;长分钟 = ((秒 % SECONDS_PER_HOUR)/SECONDS_PER_MINUTE);long secs = (seconds % SECONDS_PER_MINUTE);返回新的长[]{小时，分钟，秒}；}}

我得到的输出是 29 年 8 个月 24 天 12 小时 0 分 50 秒.我已经从这个(结果:29年8个月24天12小时0分钟和 50 秒).

更新

由于我从两个不同的站点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);

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

从此link 它应该是 29 年 8 个月 24 天 22 小时 54 分 50 秒.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需要有正确的结果.

解决方案

不幸的是，似乎没有跨时间的 period 类，因此您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，可以在一定程度上简化它.这是一种计算差异的方法，虽然不一定是最快的:

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 );长年 = tempDateTime.until( toDateTime, ChronoUnit.YEARS );tempDateTime = tempDateTime.plusYears(years);长月 = 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 );长分钟 = tempDateTime.until( toDateTime, ChronoUnit.MINUTES );tempDateTime = tempDateTime.plusMinutes( 分钟);long seconds = tempDateTime.until( toDateTime, ChronoUnit.SECONDS );System.out.println(年+年"+月 + 月" +天 + 天" +小时 + " 小时 " +分钟 + "分钟" +秒 + "秒");//打印:29 年 8 个月 24 天 22 小时 54 分 50 秒.

基本思想是:创建一个临时的开始日期并获得完整的年份结束.然后按年数调整该日期，使开始日期距结束日期不到一年.按降序对每个时间单位重复此操作.

最后免责声明: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/检查夏令时或其他变化日历(如萨摩亚的时区变化)会影响此计算.所以请谨慎使用.

I am trying to calculate the difference between two LocalDateTime.

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};
    }
}

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:

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

Update

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).

Update

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);

Then the output is coming: 29 years 8 months 25 days -1 hours -5 minutes -10 seconds.

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.

Note the question is not about which site gave me what result, I need to know the right algorithm and need to have right results.

解决方案

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.

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.

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.

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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