用java计算日期/时差[重复]

This question already has an answer here:

这个问题已经有了答案:

• How to convert Milliseconds to “X mins, x seconds” in Java? 24 answers
• 在Java中如何将毫秒转换为“X分钟，X秒”?24日答案

I want to calculate difference between 2 dates in hours/minutes/seconds.

我想用小时/分钟/秒来计算两次约会之间的差异。

I have a slight problem with my code here it is :

我的代码有点问题，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."); 

This should produce :

这将产生:

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

However I get this result :

但是我得到的结果是:

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

Can anyone see what I'm doing wrong here ?

有人能看出我做错了什么吗?

17 个解决方案

#1

try

试一试

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

#2

I would prefer to use suggested java.util.concurrent.TimeUnit class.

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); 

#3

If you are able to use external libraries I would recommend you to use Joda-Time, noting that:

如果您能够使用外部库，我建议您使用Joda-Time，注意:

Joda-Time is the de facto standard date and time library for Java prior to Java SE 8. Users are now asked to migrate to java.time (JSR-310).

Joda-Time是Java SE 8之前的实际标准日期和时间库。现在要求用户迁移到java。时间(jsr - 310)。

Example for between calculation:

计算之间的例子:

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

#4

Since Java 5, you can use java.util.concurrent.TimeUnit to avoid the use of Magic Numbers like 1000 and 60 in your code.

由于Java 5，您可以使用Java .util.concurrent。避免在代码中使用像1000和60这样的神奇数字的时间单位。

By the way, you should take care to leap seconds in your computation: the last minute of a year may have an additional leap second so it indeed lasts 61 seconds instead of expected 60 seconds. The ISO specification even plan for possibly 61 seconds. You can find detail in java.util.Date javadoc.

顺便说一下，你应该在计算中注意闰秒:一年中的最后一分钟可能会有额外的闰秒，所以它实际上会持续61秒，而不是预期的60秒。ISO规范甚至可能规划61秒。您可以在java.util中找到细节。日期javadoc。

#5

Try this for a friendly representation of time differences (in milliseconds):

对于时间差异的友好表示(以毫秒为单位)，请尝试以下方法:

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";
    }
}

#6

This is more of a maths problem than a java problem basically.

这基本上是一个数学问题而不是java问题。

The result you receive is correct. This because 225 seconds is 3 minutes (when doing an integral division). What you want is the this:

你收到的结果是正确的。因为225秒是3分钟(做积分除法时)。你想要的是:

• divide by 1000 to get the number of seconds -> rest is milliseconds
• 除以1000得到秒数->休息是毫秒
• divide that by 60 to get number of minutes -> rest are seconds
• 除以60得到分钟数->休息是秒
• divide that by 60 to get number of hours -> rest are minutes
• 除以60得到小时->休息是分钟

or in java:

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

difference-between-two-dates-in-java

difference-between-two-dates-in-java

Extracted the code from the link

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;
    }
}

#8

Create a Date object using the diffence between your times as a constructor,
then use Calendar methods to get values ..

使用时间之间的差异作为构造函数创建一个日期对象，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);

#9

// 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.");

#10

Joda-Time

Joda-Time 2.3 library offers already-debugged code for this chore.

Joda-Time 2.3库为这个任务提供了已经调试好的代码。

Joad-Time includes three classes to represent a span of time: Period, Interval, and Duration. Period tracks a span as a number of months, days, hours, etc. (not tied to the timeline).

joadtime包含三个类来表示时间跨度:周期、间隔和持续时间。期间跟踪一段时间，包括几个月、几天、几个小时等(不受时间限制)。

// © 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 );

When run…

运行时……

output: 3 minutes and 45 seconds

#11

I know this is an old question, but I ended up doing something slightly different from the accepted answer. People talk about the TimeUnit class, but there were no answers using this in the way OP wanted it.

我知道这是一个老问题，但我最终做了一些与公认答案稍有不同的事情。人们谈论的是TimeUnit类，但是在OP想要它的方式上没有答案。

So here's another solution, should someone come by missing it ;-)

所以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);
    }
}

Although just calculating the difference yourself can be done, it's not very meaningful to do it like that and I think TimeUnit is a highly overlooked class.

虽然仅仅计算自己可以做的不同，但是这样做并没有什么意义，我认为TimeUnit是一个被高度忽略的类。

#12

Here is a suggestion, using TimeUnit, to obtain each time part and format them.

这里有一个建议，使用时间单元，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));

The result is: 01:06:15,928

结果是:01:06:15,928

#13

Here is my code.

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

long diffSeconds = (diff / 1000)%60;
try this and let me know if it works correctly...

长扩散率= (diff / 1000)%60;试试这个，让我知道它是否正确……

#15

Well, I'll try yet another code sample:

那么，我再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

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);

You can use Apache Commons Duration Format Utils. It formats like SimpleDateFormatter

您可以使用Apache Commons Duration格式Utils。这格式像simpledateformat

Output:

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

输出:0天0小时0分钟1秒

#17

As said before - think this is a good answer

如前所述，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;
 }