我希望以小时/分钟/秒计算两个日期之间的差异。
我的代码存在轻微问题：
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？
答案 0 :(得分：209)
我更愿意使用建议的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);
答案 1 :(得分：93)
试
long diffSeconds = diff / 1000 % 60;
long diffMinutes = diff / (60 * 1000) % 60;
long diffHours = diff / (60 * 60 * 1000);
注意：这假定diff是非负面的。
答案 2 :(得分：39)
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的事实标准日期和时间库。现在要求用户迁移到java.time（JSR-310）。
计算之间的示例：
Seconds.between(startDate, endDate);
Days.between(startDate, endDate);
答案 3 :(得分：18)
从Java 5开始，您可以使用java.util.concurrent.TimeUnit来避免在代码中使用像Magic和1000这样的Magic Numbers。
顺便说一句，你应该注意计算中的闰秒：一年的最后一分钟可能有一个额外的闰秒，所以它确实持续61秒而不是预期的60秒。 ISO规范甚至计划可能61秒。您可以在java.util.Date javadoc。
答案 4 :(得分：12)
尝试使用友好的时差表示（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";
}
}
答案 5 :(得分：6)
这基本上是一个数学问题，而不是java问题。
您收到的结果是正确的。这是因为225秒是3分钟（进行积分时）。你想要的是这个：
或在java中：
int millis = diff % 1000;
diff/=1000;
int seconds = diff % 60;
diff/=60;
int minutes = diff % 60;
diff/=60;
hours = diff;
答案 6 :(得分：3)
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;
}
}
答案 7 :(得分：3)
我知道这是一个老问题，但我最终做的事情与接受的答案略有不同。人们谈论TimeUnit课程，但是在OP想要的方式中没有使用这个答案。
所以这是另一个解决方案，如果有人错过了它，那么; - ）
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异，但这样做并不是很有意义，我认为TimeUnit是一个被高度忽视的课程。
答案 8 :(得分：3)
以下建议使用TimeUnit获取每个时间部分并对其进行格式化。
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
答案 9 :(得分：2)
使用您的时间之间的差异作为构造函数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);
答案 10 :(得分：2)
// 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.");
答案 11 :(得分：2)
Joda-Time 2.3库提供了已经调试过的家务代码。
Joad-Time包含三个代表一段时间的课程：Period，Interval和Duration。 Period跟踪跨度为月，日，小时等数量（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 );
跑步时......
output: 3 minutes and 45 seconds
答案 12 :(得分：1)
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;
}
}
答案 13 :(得分：0)
long diffSeconds =（diff / 1000）％60;
尝试这个，让我知道它是否正常工作......
答案 14 :(得分：0)
好吧，我会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;
}
答案 15 :(得分：0)
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)
答案 16 :(得分：0)
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;
}