我想以小时/分钟/秒为单位计算两个日期之间的时差。

我的代码在这里有一个小问题,它是:

String dateStart = "11/03/14 09:29:58";
String dateStop = "11/03/14 09:33:43";

// Custom date format
SimpleDateFormat format = new SimpleDateFormat("yy/MM/dd HH:mm:ss");  

Date d1 = null;
Date d2 = null;
try {
    d1 = format.parse(dateStart);
    d2 = format.parse(dateStop);
} catch (ParseException e) {
    e.printStackTrace();
}    

// Get msec from each, and subtract.
long diff = d2.getTime() - d1.getTime();
long diffSeconds = diff / 1000;         
long diffMinutes = diff / (60 * 1000);         
long diffHours = diff / (60 * 60 * 1000);                      
System.out.println("Time in seconds: " + diffSeconds + " seconds.");         
System.out.println("Time in minutes: " + diffMinutes + " minutes.");         
System.out.println("Time in hours: " + diffHours + " hours."); 


这应该产生:

Time in seconds: 45 seconds.
Time in minutes: 3 minutes.
Time in hours: 0 hours.


但是我得到这个结果:

Time in seconds: 225 seconds.
Time in minutes: 3 minutes.
Time in hours: 0 hours.


在这里有人能看到我在做什么错吗?

评论

发现了类似的问题:stackoverflow.com/questions/625433/…

有一种更好的方法:stackoverflow.com/a/15541322/562769

对于Java 8+,有一种真正简单的方法:stackoverflow.com/a/23176621/1548776

#1 楼

请尝试

long diffSeconds = diff / 1000 % 60;  
long diffMinutes = diff / (60 * 1000) % 60; 
long diffHours = diff / (60 * 60 * 1000);


注意:这假设diff是非负的。

评论


@ vels4j我假设我刚刚添加的原始代码的第三行。感谢您指出这一点。

– Peter Lawrey
18年3月3日,11:43



但是diffMinutes仍然是错误的

–vels4j
18年3月3日,12:10

@ vels4j用什么方式?

– Peter Lawrey
18年3月3日在15:32

我已经更新了答案,请检查一下。澄清后放弃我的编辑

–vels4j
18年11月8日在6:05

@ vels4j您说的是错误的,因为6似乎是正确的答案,因为您在OPs问题中也​​有几个小时。

– Peter Lawrey
18年11月8日在9:47

#2 楼

我宁愿使用建议的java.util.concurrent.TimeUnit类。

long diff = d2.getTime() - d1.getTime();//as given

long seconds = TimeUnit.MILLISECONDS.toSeconds(diff);
long minutes = TimeUnit.MILLISECONDS.toMinutes(diff); 


评论


@马克我不同意。您错过了问题的重点。尽管在帖子的其余部分中使用了“以秒为单位的时间”这一短语,但他明确表示实际上并不需要简单的转换,但他需要余数。这比接受的答案(使用方法调用,即使在JVM字节码中也包含一些指令)的效率低,不清楚(它更长,并且坦率地说,如果在此上下文中发现“ 1000”或“ 60”是幻数),则效率较低。 ,他们无法使用完整的平台),并且,从根本上讲,它无法满足OP的要求。

–Parthian Shot
2014年8月8日在17:25

有一种方法可以为此编写通用方法。见stackoverflow.com/a/10650881/82609

–塞巴斯蒂安·洛伯(Sebastien Lorber)
16年1月29日在10:20

如果我想要,相差数周,数月,数年?

– jose920405
2016年9月15日16:51

@ jose920405在这种情况下,您需要乔达时间

–约翰·亨克尔(John Henckel)
17年6月29日在21:13

这将给OP指定他们不想要的答案。

– Peter Lawrey
18年3月3日在15:33

#3 楼

如果您能够使用外部库,则建议您使用Joda-Time,请注意:


Joda-Time是Java之前的Java的事实上的标准日期和时间库。 SE 8.现在,要求用户迁移到java.time(JSR-310)。


计算示例:

Seconds.between(startDate, endDate);
Days.between(startDate, endDate);


评论


请注意,如果要处理2个java.util.Date对象,则需要使用Days.daysBetween(LocalDate.fromDateFields(startDate),LocalDate.fromDateFields(endDate)));

–user799188
2012年10月9日3:00



我创建了一个如何使用乔达时间的示例:stackoverflow.com/a/15541322/562769

–马丁·托马
13年3月21日在7:53

目前的做法是什么? (使用JSR-310)

–static_rtti
18 Mar 29 '18 at 10:27

#4 楼

从Java 5开始,您可以使用java.util.concurrent.TimeUnit避免在代码中使用诸如1000和60之类的幻数。

顺便说一句,在计算时应格外注意:一年可能还会有额外的leap秒,因此实际上持续了61秒,而不是预期的60秒。 ISO规范甚至可能计划61秒钟。您可以在java.util.Date javadoc中找到详细信息。

评论


除非有充分的商业理由将那些飞跃的秒数包括在内,否则您会很安全地将它们视为有趣但又不重要的科学好奇心。

–食指
13年4月16日在2:44

我同意“ le秒”是个小技巧。但是,夏时制或时区差异又如何呢?

–伊夫·马丁(Yves Martin)
13年5月3日在10:28

@YvesMartin“时区差异”看看他的日期格式。没有时区的地方。如果他确实希望跨时区工作,那么他将面临更大的问题。但是,夏令时是一个问题。

–Parthian Shot
2014年8月8日在17:29

«它的确持续60秒,而不是预期的59秒»您的意思是它持续61秒(0..60),而不是预期的60秒(0..59)。 ;)

– Andrea Lazzarotto
16年5月24日在11:45

修正了我的评论...一个经典的纠察队/间距问题!感谢您指出

–伊夫·马丁(Yves Martin)
16年5月24日15:15

#5 楼

尝试以下操作以方便地表示时差(以毫秒为单位):

String friendlyTimeDiff(long timeDifferenceMilliseconds) {
    long diffSeconds = timeDifferenceMilliseconds / 1000;
    long diffMinutes = timeDifferenceMilliseconds / (60 * 1000);
    long diffHours = timeDifferenceMilliseconds / (60 * 60 * 1000);
    long diffDays = timeDifferenceMilliseconds / (60 * 60 * 1000 * 24);
    long diffWeeks = timeDifferenceMilliseconds / (60 * 60 * 1000 * 24 * 7);
    long diffMonths = (long) (timeDifferenceMilliseconds / (60 * 60 * 1000 * 24 * 30.41666666));
    long diffYears = timeDifferenceMilliseconds / ((long)60 * 60 * 1000 * 24 * 365);

    if (diffSeconds < 1) {
        return "less than a second";
    } else if (diffMinutes < 1) {
        return diffSeconds + " seconds";
    } else if (diffHours < 1) {
        return diffMinutes + " minutes";
    } else if (diffDays < 1) {
        return diffHours + " hours";
    } else if (diffWeeks < 1) {
        return diffDays + " days";
    } else if (diffMonths < 1) {
        return diffWeeks + " weeks";
    } else if (diffYears < 1) {
        return diffMonths + " months";
    } else {
        return diffYears + " years";
    }
}


评论


在线上long diffYears =(long)(timeDifferenceMilliseconds /(60 * 60 * 1000 * 24 * 365));您需要稍后进行长时间转换,否则除数为int并溢出:diffYears =(timeDifferenceMilliseconds /((long)60 * 60 * 1000 * 24 * 365));

– Pavel Niedoba
16年4月14日在13:38

我已经修复了代码

– Pavel Niedoba
16年4月14日在14:13

#6 楼

基本上,这比Java问题更多是数学问题。

您收到的结果是正确的。这是因为225秒是3分钟(进行整数除法时)。您想要的是这样的:


除以1000得到秒数->休息是毫秒
除以60得到分钟数->休息是seconds
用60除以得到小时数->休息是分钟

或在java中:

int millis = diff % 1000;
diff/=1000;
int seconds = diff % 60;
diff/=60;
int minutes = diff % 60;
diff/=60;
hours = diff;


#7 楼

这是一个建议,使用TimeUnit来获取每个时间部分并将其格式化。

private static String formatDuration(long duration) {
    long hours = TimeUnit.MILLISECONDS.toHours(duration);
    long minutes = TimeUnit.MILLISECONDS.toMinutes(duration) % 60;
    long seconds = TimeUnit.MILLISECONDS.toSeconds(duration) % 60;
    long milliseconds = duration % 1000;
    return String.format("%02d:%02d:%02d,%03d", hours, minutes, seconds, milliseconds);
}

SimpleDateFormat sdf = new SimpleDateFormat("HH:mm:ss,SSS");
Date startTime = sdf.parse("01:00:22,427");
Date now = sdf.parse("02:06:38,355");
long duration = now.getTime() - startTime.getTime();
System.out.println(formatDuration(duration));


结果是:01:06:15,928

评论


谢谢你的作品

–菲利普
8月21日3:06

#8 楼

使用您的时间差作为构造函数创建一个Date对象,
然后使用Calendar方法获取值..

Date diff = new Date(d2.getTime() - d1.getTime());

Calendar calendar = Calendar.getInstance();
calendar.setTime(diff);
int hours = calendar.get(Calendar.HOUR_OF_DAY);
int minutes = calendar.get(Calendar.MINUTE);
int seconds = calendar.get(Calendar.SECOND);


评论


这些方法已被弃用。

– djechlin
13年2月14日在23:15

工作正常,但您一定不要忘记设置TimeZone:Calendar calendar = Calendar.getInstance(TimeZone.getTimeZone(“ UTC”));

– Amra
13年7月11日在13:19

#9 楼

我知道这是一个老问题,但是我最终做了一些与接受的答案略有不同的事情。人们谈论TimeUnit类,但是以OP想要的方式使用它没有任何答案。

因此,这是另一个解决方案,如果有人错过了它;-)

public class DateTesting {
    public static void main(String[] args) {
        String dateStart = "11/03/14 09:29:58";
        String dateStop = "11/03/14 09:33:43";

        // Custom date format
        SimpleDateFormat format = new SimpleDateFormat("yy/MM/dd HH:mm:ss");  

        Date d1 = null;
        Date d2 = null;
        try {
            d1 = format.parse(dateStart);
            d2 = format.parse(dateStop);
        } catch (ParseException e) {
            e.printStackTrace();
        }    

        // Get msec from each, and subtract.
        long diff = d2.getTime() - d1.getTime();

        long days = TimeUnit.MILLISECONDS.toDays(diff);
        long remainingHoursInMillis = diff - TimeUnit.DAYS.toMillis(days);
        long hours = TimeUnit.MILLISECONDS.toHours(remainingHoursInMillis);
        long remainingMinutesInMillis = remainingHoursInMillis - TimeUnit.HOURS.toMillis(hours);
        long minutes = TimeUnit.MILLISECONDS.toMinutes(remainingMinutesInMillis);
        long remainingSecondsInMillis = remainingMinutesInMillis - TimeUnit.MINUTES.toMillis(minutes);
        long seconds = TimeUnit.MILLISECONDS.toSeconds(remainingSecondsInMillis);

        System.out.println("Days: " + days + ", hours: " + hours + ", minutes: " + minutes + ", seconds: " + seconds);
    }
}


尽管可以自己计算出差异,但这样做并不是很有意义,我认为TimeUnit是一个被人们高度忽视的类。

评论


谢谢!我以同样的方式思考,但是您加快了速度:)

– Akbolat SSS
19年11月23日在16:56

#10 楼

Java中两个日期之间的差异

从链接中提取代码

public class TimeDiff {
    /**
     * (For testing purposes)
     *
     */
    public static void main(String[] args) {
        Date d1 = new Date();
        try { Thread.sleep(750); } catch(InterruptedException e) { /* ignore */ }      
        Date d0 = new Date(System.currentTimeMillis() - (1000*60*60*24*3)); // About 3 days ago
        long[] diff = TimeDiff.getTimeDifference(d0, d1);

        System.out.printf("Time difference is %d day(s), %d hour(s), %d minute(s), %d second(s) and %d millisecond(s)\n",
                diff[0], diff[1], diff[2], diff[3], diff[4]);
        System.out.printf("Just the number of days = %d\n",
                TimeDiff.getTimeDifference(d0, d1, TimeDiff.TimeField.DAY));
    }

    /**
     * Calculate the absolute difference between two Date without
     * regard for time offsets
     *
     * @param d1 Date one
     * @param d2 Date two
     * @param field The field we're interested in out of
     * day, hour, minute, second, millisecond
     *
     * @return The value of the required field
     */
    public static long getTimeDifference(Date d1, Date d2, TimeField field) {
        return TimeDiff.getTimeDifference(d1, d2)[field.ordinal()];
    }

    /**
     * Calculate the absolute difference between two Date without
     * regard for time offsets
     *
     * @param d1 Date one
     * @param d2 Date two
     * @return The fields day, hour, minute, second and millisecond
     */
    public static long[] getTimeDifference(Date d1, Date d2) {
        long[] result = new long[5];
        Calendar cal = Calendar.getInstance();
        cal.setTimeZone(TimeZone.getTimeZone("UTC"));
        cal.setTime(d1);

        long t1 = cal.getTimeInMillis();
        cal.setTime(d2);

        long diff = Math.abs(cal.getTimeInMillis() - t1);
        final int ONE_DAY = 1000 * 60 * 60 * 24;
        final int ONE_HOUR = ONE_DAY / 24;
        final int ONE_MINUTE = ONE_HOUR / 60;
        final int ONE_SECOND = ONE_MINUTE / 60;

        long d = diff / ONE_DAY;
        diff %= ONE_DAY;

        long h = diff / ONE_HOUR;
        diff %= ONE_HOUR;

        long m = diff / ONE_MINUTE;
        diff %= ONE_MINUTE;

        long s = diff / ONE_SECOND;
        long ms = diff % ONE_SECOND;
        result[0] = d;
        result[1] = h;
        result[2] = m;
        result[3] = s;
        result[4] = ms;

        return result;
    }

    public static void printDiffs(long[] diffs) {
        System.out.printf("Days:         %3d\n", diffs[0]);
        System.out.printf("Hours:        %3d\n", diffs[1]);
        System.out.printf("Minutes:      %3d\n", diffs[2]);
        System.out.printf("Seconds:      %3d\n", diffs[3]);
        System.out.printf("Milliseconds: %3d\n", diffs[4]);
    }

    public static enum TimeField {DAY,
        HOUR,
        MINUTE,
        SECOND,
        MILLISECOND;
    }
}


评论


好。时间偏移量很重要-即使根据相同的偏移量计算出差异,但两个“日历日期”之间的任何夏时制差异也会对结果产生影响。您应该使用java.util.concurrent.TimeUnit作为常量。

–伊夫·马丁(Yves Martin)
13年5月3日在10:26



#11 楼

// d1, d2 are dates
long diff = d2.getTime() - d1.getTime();

long diffSeconds = diff / 1000 % 60;
long diffMinutes = diff / (60 * 1000) % 60;
long diffHours = diff / (60 * 60 * 1000) % 24;
long diffDays = diff / (24 * 60 * 60 * 1000);

System.out.print(diffDays + " days, ");
System.out.print(diffHours + " hours, ");
System.out.print(diffMinutes + " minutes, ");
System.out.print(diffSeconds + " seconds.");


#12 楼

Joda-Time

Joda-Time 2.3库提供了已经调试好的此杂物的代码。

Joad-Time包括三个代表时间跨度的类:PeriodInterval,和DurationPeriod跟踪跨度为月,日,小时等(不与时间轴相关)。

 // © 2013 Basil Bourque. This source code may be used freely forever by anyone taking full responsibility for doing so.

// Specify a time zone rather than rely on default.
// Necessary to handle Daylight Saving Time (DST) and other anomalies.
DateTimeZone timeZone = DateTimeZone.forID( "America/Montreal" );

DateTimeFormatter formatter = DateTimeFormat.forPattern( "yy/MM/dd HH:mm:ss" ).withZone( timeZone ); 

DateTime dateTimeStart = formatter.parseDateTime( "11/03/14 09:29:58" );
DateTime dateTimeStop = formatter.parseDateTime( "11/03/14 09:33:43" );
Period period = new Period( dateTimeStart, dateTimeStop );

PeriodFormatter periodFormatter = PeriodFormat.getDefault();
String output = periodFormatter.print( period );

System.out.println( "output: " + output );
 


运行时…

 output: 3 minutes and 45 seconds
 


#13 楼

这是我的代码。

import java.util.Date;

// to calculate difference between two days
public class DateDifference {

// to calculate difference between two dates in milliseconds
public long getDateDiffInMsec(Date da, Date db) {
    long diffMSec = 0;
    diffMSec = db.getTime() - da.getTime();
    return diffMSec;
}

// to convert Milliseconds into DD HH:MM:SS format.
public String getDateFromMsec(long diffMSec) {
    int left = 0;
    int ss = 0;
    int mm = 0;
    int hh = 0;
    int dd = 0;
    left = (int) (diffMSec / 1000);
    ss = left % 60;
    left = (int) left / 60;
    if (left > 0) {
        mm = left % 60;
        left = (int) left / 60;
        if (left > 0) {
            hh = left % 24;
            left = (int) left / 24;
            if (left > 0) {
                dd = left;
            }
        }
    }
    String diff = Integer.toString(dd) + " " + Integer.toString(hh) + ":"
            + Integer.toString(mm) + ":" + Integer.toString(ss);
    return diff;

}
}


#14 楼

long diffSeconds =(diff / 1000)%60;
尝试一下,让我知道它是否正确运行...

#15 楼

好吧,我将尝试另一个代码示例:

/**
 * Calculates the number of FULL days between to dates
 * @param startDate must be before endDate
 * @param endDate must be after startDate
 * @return number of day between startDate and endDate
 */
public static int daysBetween(Calendar startDate, Calendar endDate) {
    long start = startDate.getTimeInMillis();
    long end = endDate.getTimeInMillis();
    // It's only approximation due to several bugs (@see java.util.Date) and different precision in Calendar chosen
    // by user (ex. day is time-quantum).
    int presumedDays = (int) TimeUnit.MILLISECONDS.toDays(end - start);
    startDate.add(Calendar.DAY_OF_MONTH, presumedDays);
    // if we still didn't reach endDate try it with the step of one day
    if (startDate.before(endDate)) {
        startDate.add(Calendar.DAY_OF_MONTH, 1);
        ++presumedDays;
    }
    // if we crossed endDate then we must go back, because the boundary day haven't completed yet
    if (startDate.after(endDate)) {
        --presumedDays;
    }
    return presumedDays;
}


#16 楼

Date startTime = new Date();
//...
//... lengthy jobs
//...
Date endTime = new Date();
long diff = endTime.getTime() - startTime.getTime();
String hrDateText = DurationFormatUtils.formatDuration(diff, "d 'day(s)' H 'hour(s)' m 'minute(s)' s 'second(s)' ");
System.out.println("Duration : " + hrDateText);


您可以使用Apache Commons Duration Format Utils。格式类似于SimpleDateFormatter

输出:0 days(s) 0 hour(s) 0 minute(s) 1 second(s)

#17 楼

如前所述-认为这是一个很好的答案

/**
 * @param d2 the later date 
 * @param d1 the earlier date
 * @param timeUnit - Example Calendar.HOUR_OF_DAY
 * @return
 */
public static int getTimeDifference(Date d2,Date d1, int timeUnit) {
     Date diff = new Date(d2.getTime() - d1.getTime());

     Calendar calendar = Calendar.getInstance();
     calendar.setTime(diff);
     int hours = calendar.get(Calendar.HOUR_OF_DAY);
     int minutes = calendar.get(Calendar.MINUTE);
     int seconds = calendar.get(Calendar.SECOND);
     if(timeUnit==Calendar.HOUR_OF_DAY)
         return hours;
     if(timeUnit==Calendar.MINUTE)
         return minutes;
     return seconds;
 }


评论


这将失败,时间单位为分钟,时间差超过一小时

– Pavel Niedoba
16-2-26在23:29