用 java 计算日期/时间差

我想以小时/分钟/秒计算 计算两个日期之间的差额。

我的代码有个小问题，是这样的:

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.");

这应该会产生:

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

然而我得到了这样的结果:

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

有人知道我哪里做错了吗？

470710 次浏览
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最佳答案

试试看

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

注意: 这里假设 diff是非负的。

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如果您能够使用外部库，我建议您使用 Joda 时间，注意:

Joda-Time 是 javase8之前的 Java 行业标准日期和时间库。现在要求用户迁移到 java.time (JSR-310)。

计算间隔的例子:

Seconds.between(startDate, endDate);
Days.between(startDate, endDate);
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长差秒 = (diff/1000)% 60;
试试这个，让我知道它是否工作正确..。

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在爪哇的两次约会之间的区别

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;
}
}
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这基本上是一个数学问题而不是 Java 问题。

你收到的结果是正确的。这是因为225秒等于3分钟(在做积分除法时)。你想要的是这个:

  • 除以1000得到秒数-> 休息时间是毫秒
  • 除以60得到分钟数-> 休息时间为秒
  • 除以60得到小时数-> 休息时间是分钟

或者用爪哇语说:

int millis = diff % 1000;
diff/=1000;
int seconds = diff % 60;
diff/=60;
int minutes = diff % 60;
diff/=60;
hours = diff;
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从 Java5开始，您可以使用 java.util.concurrent.TimeUnit来避免在代码中使用魔术数字，比如1000和60。

顺便说一下，在计算中应该注意闰秒: 一年的最后一分钟可能有一个额外的闰秒，因此它确实持续了61秒，而不是预期的60秒。ISO 规范甚至可能计划61秒。您可以在 java.util.Date javadoc 中找到详细信息。

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这是我的密码。

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;


}
}
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使用作为构造函数的时间差创建一个 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);
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请尝试以下方式友好地表示时间差异(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";
}
}
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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);
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好的，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;
}
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// 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.");
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Joda 时间

Joda-Time 2.3库为这个琐事提供了已经调试好的代码。

Joad-Time 包括三个表示时间跨度的类: Period、 Interval和 Duration。Period以月、日、小时等数字来跟踪跨度(不与时间线相关联)。

// © 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 );

跑的时候..。

output: 3 minutes and 45 seconds
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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);


你可以使用 ApacheCommons 持续时间格式实用程序。它的格式类似于 SimpleDateFormatter

产出:

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

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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;
}
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我知道这是一个老问题，但我最终做了一些与公认答案略有不同的事情。人们谈论的 TimeUnit类，但没有答案使用这一方式 OP 想要它。

所以这里有另一个解决方案，如果有人来错过了它; -)

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);
}
}

虽然只是计算差异自己可以做，这样做没有什么意义，我认为 TimeUnit是一个高度忽视类。

小开

这里是一个建议，使用 TimeUnit，获取每个时间部分和格式化他们。

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


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