在java中计算日期/时间差[重复]

Question

我想以小时/分钟/秒计算两个日期之间的差异。

我的代码有点小问题：

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.

但是我得到了这个结果：

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

谁能看到我在这里做错了什么？

Answer 1

我更喜欢使用建议的 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); 

Answer 2

试试

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

注意：这里假设 diff 是非负数。

Answer 3

如果您能够使用外部库，我建议您使用Joda-Time，注意：

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

计算间隔示例：

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

Answer 4

从 Java 5 开始，您可以使用 java.util.concurrent.TimeUnit 来避免在代码中使用像 1000 和 60 这样的幻数。

顺便说一句，您应该注意计算闰秒：一年的最后一分钟可能会有一个额外的闰秒，因此它确实持续了 61 秒，而不是预期的 60 秒。 ISO 规范甚至计划可能为 61 秒。您可以在java.util.Datejavadoc 中找到详细信息。

Answer 5

试试这个以友好地表示时间差（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";
    }
}

Answer 6

这基本上是一个数学问题，而不是一个 java 问题。

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

• 除以 1000 得到秒数 -> 其余为毫秒
• 除以 60 得到分钟数 -> 其余为秒
• 除以 60 得到小时数 -> 休息是分钟

或者在java中：

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

Answer 7

这是一个建议，使用TimeUnit，获取每个时间部分并格式化它们。

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

Answer 8

我知道这是一个老问题，但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，但没有以 OP 想要的方式使用它的答案。

所以这是另一种解决方案，如果有人错过它;-)

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);
    }
}

虽然只是自己计算差值也可以，但那样做并没有多大意义，而且我认为TimeUnit是一个被高度忽视的类。

Answer 9

使用您作为构造函数的时间差异创建一个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);

Answer 10

difference-between-two-dates-in-java

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;
    }
}

Answer 11

// 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.");

Answer 12

乔达时间

Joda-Time 2.3 库为这项杂务提供了已经调试过的代码。

Joad-Time 包括三个类来表示时间跨度：Period、Interval 和 Duration。 Period 将跨度跟踪为月数、天数、小时数等（与时间线无关）。

// © 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

Answer 13

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;

}
}

Answer 14

long diffSeconds = (diff / 1000)%60;
试试这个，让我知道它是否正常工作......

Answer 15

好吧，我再试试另一个代码示例：

/**
 * 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;
}

Answer 16

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。它的格式类似于SimpleDateFormatter

输出：

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

Answer 17

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;
 }