How to find the duration of difference between two dates in java?

I have two objects of DateTime, which need to find the duration of their difference,

I have the following code but not sure how to continue it to get to the expected results as following:

Example:

      11/03/14 09:30:58
11/03/14 09:33:43
elapsed time is 02 minutes and 45 seconds
-----------------------------------------------------
11/03/14 09:30:58
11/03/15 09:30:58
elapsed time is a day
-----------------------------------------------------
11/03/14 09:30:58
11/03/16 09:30:58
elapsed time is two days
-----------------------------------------------------
11/03/14 09:30:58
11/03/16 09:35:58
elapsed time is two days and 05 minutes
      

Code:

    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 % 60;
long diffMinutes = diff / (60 * 1000) % 60;
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.");
424186 次浏览

使用 Joda 时间

DateTime startTime, endTime;
Period p = new Period(startTime, endTime);
long hours = p.getHours();
long minutes = p.getMinutes();

Joda Time 有一个关于时间的概念间隔:

Interval interval = new Interval(oldTime, new Instant());

再举个例子 日期差异

One more 林克

或者使用 Java-8(它集成了 Joda-Time 概念)

Instant start, end;//
Duration dur = Duration.between(start, stop);
long hours = dur.toHours();
long minutes = dur.toMinutes();

正如迈克尔•博格沃特(Michael Borgwardt)在 his answer here中写道:

int diffInDays = (int)( (newerDate.getTime() - olderDate.getTime())
/ (1000 * 60 * 60 * 24) )

Note that this works with UTC dates, so the difference may be a day 如果你看看当地的日期。并得到它的工作正确与 当地的日期需要一个完全不同的方法,由于白天 节省时间。

使用 Java 内置类 时间单位可以更好地处理日期差异转换。它提供了这样做的实用方法:

Date startDate = // Set start date
Date endDate   = // Set end date


long duration  = endDate.getTime() - startDate.getTime();


long diffInSeconds = TimeUnit.MILLISECONDS.toSeconds(duration);
long diffInMinutes = TimeUnit.MILLISECONDS.toMinutes(duration);
long diffInHours = TimeUnit.MILLISECONDS.toHours(duration);
long diffInDays = TimeUnit.MILLISECONDS.toDays(duration);

试试以下方法

{
Date dt2 = new DateAndTime().getCurrentDateTime();


long diff = dt2.getTime() - dt1.getTime();
long diffSeconds = diff / 1000 % 60;
long diffMinutes = diff / (60 * 1000) % 60;
long diffHours = diff / (60 * 60 * 1000);
int diffInDays = (int) ((dt2.getTime() - dt1.getTime()) / (1000 * 60 * 60 * 24));


if (diffInDays > 1) {
System.err.println("Difference in number of days (2) : " + diffInDays);
return false;
} else if (diffHours > 24) {


System.err.println(">24");
return false;
} else if ((diffHours == 24) && (diffMinutes >= 1)) {
System.err.println("minutes");
return false;
}
return true;
}
Date d2 = new Date();
Date d1 = new Date(1384831803875l);


long diff = d2.getTime() - d1.getTime();
long diffSeconds = diff / 1000 % 60;
long diffMinutes = diff / (60 * 1000) % 60;
long diffHours = diff / (60 * 60 * 1000);
int diffInDays = (int) diff / (1000 * 60 * 60 * 24);


System.out.println(diffInDays+"  days");
System.out.println(diffHours+"  Hour");
System.out.println(diffMinutes+"  min");
System.out.println(diffSeconds+"  sec");

下面是如何在 Java8中解决这个问题,就像 shaimz 给出的答案一样。

资料来源: http://docs.oracle.com/javase/tutorial/datetime/iso/period.html

LocalDate today = LocalDate.now();
LocalDate birthday = LocalDate.of(1960, Month.JANUARY, 1);


Period p = Period.between(birthday, today);
long p2 = ChronoUnit.DAYS.between(birthday, today);


System.out.println("You are " + p.getYears() + " years, " + p.getMonths() + " months, and " + p.getDays() + " days old. (" + p2 + " days total)");

该守则产生的输出类似于以下内容:

You are 53 years, 4 months, and 29 days old. (19508 days total)

我们必须使用 LocalDateTimehttp://docs.oracle.com/javase/8/docs/api/java/time/LocalDateTime.html来获得小时、分钟和秒的差异。

您可以创建如下方法

public long getDaysBetweenDates(Date d1, Date d2){
return TimeUnit.MILLISECONDS.toDays(d1.getTime() - d2.getTime());
}

此方法将返回两天之间的天数。

In Java 8, you can make of DateTimeFormatter, Duration, and LocalDateTime. Here is an example:

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


final DateTimeFormatter formatter = new DateTimeFormatterBuilder()
.appendValue(ChronoField.MONTH_OF_YEAR, 2)
.appendLiteral('/')
.appendValue(ChronoField.DAY_OF_MONTH, 2)
.appendLiteral('/')
.appendValueReduced(ChronoField.YEAR, 2, 2, 2000)
.appendLiteral(' ')
.appendValue(ChronoField.HOUR_OF_DAY, 2)
.appendLiteral(':')
.appendValue(ChronoField.MINUTE_OF_HOUR, 2)
.appendLiteral(':')
.appendValue(ChronoField.SECOND_OF_MINUTE, 2)
.toFormatter();


final LocalDateTime start = LocalDateTime.parse(dateStart, formatter);
final LocalDateTime stop = LocalDateTime.parse(dateStop, formatter);


final Duration between = Duration.between(start, stop);


System.out.println(start);
System.out.println(stop);
System.out.println(formatter.format(start));
System.out.println(formatter.format(stop));
System.out.println(between);
System.out.println(between.get(ChronoUnit.SECONDS));

这是密码:

        String date1 = "07/15/2013";
String time1 = "11:00:01";
String date2 = "07/16/2013";
String time2 = "22:15:10";
String format = "MM/dd/yyyy HH:mm:ss";
SimpleDateFormat sdf = new SimpleDateFormat(format);
Date fromDate = sdf.parse(date1 + " " + time1);
Date toDate = sdf.parse(date2 + " " + time2);


long diff = toDate.getTime() - fromDate.getTime();
String dateFormat="duration: ";
int diffDays = (int) (diff / (24 * 60 * 60 * 1000));
if(diffDays>0){
dateFormat+=diffDays+" day ";
}
diff -= diffDays * (24 * 60 * 60 * 1000);


int diffhours = (int) (diff / (60 * 60 * 1000));
if(diffhours>0){
dateFormat+=diffhours+" hour ";
}
diff -= diffhours * (60 * 60 * 1000);


int diffmin = (int) (diff / (60 * 1000));
if(diffmin>0){
dateFormat+=diffmin+" min ";
}
diff -= diffmin * (60 * 1000);


int diffsec = (int) (diff / (1000));
if(diffsec>0){
dateFormat+=diffsec+" sec";
}
System.out.println(dateFormat);

结果是:

duration: 1 day 11 hour 15 min 9 sec

这是我写的一个程序,它获取两个日期之间的天数(这里没有时间)。

import java.util.Scanner;
public class HelloWorld {
public static void main(String args[]) {
Scanner s = new Scanner(System.in);
System.out.print("Enter starting date separated by dots: ");
String inp1 = s.nextLine();
System.out.print("Enter ending date separated by dots: ");
String inp2 = s.nextLine();
int[] nodim = {
0,
31,
28,
31,
30,
31,
30,
31,
31,
30,
31,
30,
31
};
String[] inpArr1 = split(inp1);
String[] inpArr2 = split(inp2);
int d1 = Integer.parseInt(inpArr1[0]);
int m1 = Integer.parseInt(inpArr1[1]);
int y1 = Integer.parseInt(inpArr1[2]);
int d2 = Integer.parseInt(inpArr2[0]);
int m2 = Integer.parseInt(inpArr2[1]);
int y2 = Integer.parseInt(inpArr2[2]);
if (y1 % 4 == 0) nodim[2] = 29;
int diff = m1 == m2 && y1 == y2 ? d2 - (d1 - 1) : (nodim[m1] - (d1 - 1));
int mm1 = m1 + 1, mm2 = m2 - 1, yy1 = y1, yy2 = y2;
for (; yy1 <= yy2; yy1++, mm1 = 1) {
mm2 = yy1 == yy2 ? (m2 - 1) : 12;
if (yy1 % 4 == 0) nodim[2] = 29;
else nodim[2] = 28;
if (mm2 == 0) {
mm2 = 12;
yy2 = yy2 - 1;
}
for (; mm1 <= mm2 && yy1 <= yy2; mm1++) diff = diff + nodim[mm1];
}
System.out.print("No. of days from " + inp1 + " to " + inp2 + " is " + diff);
}
public static String[] split(String s) {
String[] retval = {
"",
"",
""
};
s = s + ".";
s = s + " ";
for (int i = 0; i <= 2; i++) {
retval[i] = s.substring(0, s.indexOf("."));
s = s.substring((s.indexOf(".") + 1), s.length());
}
return retval;
}
}

Http://pastebin.com/hrsjttuf

   // calculating the difference b/w startDate and endDate
String startDate = "01-01-2016";
String endDate = simpleDateFormat.format(currentDate);


date1 = simpleDateFormat.parse(startDate);
date2 = simpleDateFormat.parse(endDate);


long getDiff = date2.getTime() - date1.getTime();


// using TimeUnit class from java.util.concurrent package
long getDaysDiff = TimeUnit.MILLISECONDS.toDays(getDiff);

如何在 Java 中计算两个日期之间的差异

我最近用一个简单的方法解决了类似的问题。

public static void main(String[] args) throws IOException, ParseException {
TimeZone utc = TimeZone.getTimeZone("UTC");
Calendar calendar = Calendar.getInstance(utc);
Date until = calendar.getTime();
calendar.add(Calendar.DAY_OF_MONTH, -7);
Date since = calendar.getTime();
long durationInSeconds  = TimeUnit.MILLISECONDS.toSeconds(until.getTime() - since.getTime());


long SECONDS_IN_A_MINUTE = 60;
long MINUTES_IN_AN_HOUR = 60;
long HOURS_IN_A_DAY = 24;
long DAYS_IN_A_MONTH = 30;
long MONTHS_IN_A_YEAR = 12;


long sec = (durationInSeconds >= SECONDS_IN_A_MINUTE) ? durationInSeconds % SECONDS_IN_A_MINUTE : durationInSeconds;
long min = (durationInSeconds /= SECONDS_IN_A_MINUTE) >= MINUTES_IN_AN_HOUR ? durationInSeconds%MINUTES_IN_AN_HOUR : durationInSeconds;
long hrs = (durationInSeconds /= MINUTES_IN_AN_HOUR) >= HOURS_IN_A_DAY ? durationInSeconds % HOURS_IN_A_DAY : durationInSeconds;
long days = (durationInSeconds /= HOURS_IN_A_DAY) >= DAYS_IN_A_MONTH ? durationInSeconds % DAYS_IN_A_MONTH : durationInSeconds;
long months = (durationInSeconds /=DAYS_IN_A_MONTH) >= MONTHS_IN_A_YEAR ? durationInSeconds % MONTHS_IN_A_YEAR : durationInSeconds;
long years = (durationInSeconds /= MONTHS_IN_A_YEAR);


String duration = getDuration(sec,min,hrs,days,months,years);
System.out.println(duration);
}
private static String getDuration(long secs, long mins, long hrs, long days, long months, long years) {
StringBuffer sb = new StringBuffer();
String EMPTY_STRING = "";
sb.append(years > 0 ? years + (years > 1 ? " years " : " year "): EMPTY_STRING);
sb.append(months > 0 ? months + (months > 1 ? " months " : " month "): EMPTY_STRING);
sb.append(days > 0 ? days + (days > 1 ? " days " : " day "): EMPTY_STRING);
sb.append(hrs > 0 ? hrs + (hrs > 1 ? " hours " : " hour "): EMPTY_STRING);
sb.append(mins > 0 ? mins + (mins > 1 ? " mins " : " min "): EMPTY_STRING);
sb.append(secs > 0 ? secs + (secs > 1 ? " secs " : " secs "): EMPTY_STRING);
sb.append("ago");
return sb.toString();
}

不出所料,它打印出来了: 7 days ago

参考 Shamim 的答案更新,这里是一个不使用任何第三方库完成任务的方法。只要复制方法和使用

public static String getDurationTimeStamp(String date) {


String timeDifference = "";


//date formatter as per the coder need
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");


//parse the string date-ti
// me to Date object
Date startDate = null;
try {
startDate = sdf.parse(date);
} catch (ParseException e) {
e.printStackTrace();
}


//end date will be the current system time to calculate the lapse time difference
//if needed, coder can add end date to whatever date
Date endDate = new Date();


System.out.println(startDate);
System.out.println(endDate);


//get the time difference in milliseconds
long duration = endDate.getTime() - startDate.getTime();


//now we calculate the differences in different time units
//this long value will be the total time difference in each unit
//i.e; total difference in seconds, total difference in minutes etc...
long diffInSeconds = TimeUnit.MILLISECONDS.toSeconds(duration);
long diffInMinutes = TimeUnit.MILLISECONDS.toMinutes(duration);
long diffInHours = TimeUnit.MILLISECONDS.toHours(duration);
long diffInDays = TimeUnit.MILLISECONDS.toDays(duration);


//now we create the time stamps depending on the value of each unit that we get
//as we do not have the unit in years,
//we will see if the days difference is more that 365 days, as 365 days = 1 year
if (diffInDays > 365) {
//we get the year in integer not in float
//ex- 791/365 = 2.167 in float but it will be 2 years in int
int year = (int) (diffInDays / 365);
timeDifference = year + " years ago";
System.out.println(year + " years ago");
}
//if days are not enough to create year then get the days
else if (diffInDays > 1) {
timeDifference = diffInDays + " days ago";
System.out.println(diffInDays + " days ago");
}
//if days value<1 then get the hours
else if (diffInHours > 1) {
timeDifference = diffInHours + " hours ago";
System.out.println(diffInHours + " hours ago");
}
//if hours value<1 then get the minutes
else if (diffInMinutes > 1) {
timeDifference = diffInMinutes + " minutes ago";
System.out.println(diffInMinutes + " minutes ago");
}
//if minutes value<1 then get the seconds
else if (diffInSeconds > 1) {
timeDifference = diffInSeconds + " seconds ago";
System.out.println(diffInSeconds + " seconds ago");
}


return timeDifference;
// that's all. Happy Coding :)
}

我可以试试这个,希望对你有帮助。有什么问题就告诉我。

Date startDate = java.util.Calendar.getInstance().getTime(); //set your start time
Date endDate = java.util.Calendar.getInstance().getTime(); // set  your end time


long duration = endDate.getTime() - startDate.getTime();




long diffInSeconds = TimeUnit.MILLISECONDS.toSeconds(duration);
long diffInMinutes = TimeUnit.MILLISECONDS.toMinutes(duration);
long diffInHours = TimeUnit.MILLISECONDS.toHours(duration);
long diffInDays = TimeUnit.MILLISECONDS.toDays(duration);


Toast.makeText(MainActivity.this, "Diff"
+ duration + diffInDays + diffInHours + diffInMinutes + diffInSeconds, Toast.LENGTH_SHORT).show(); **// Toast message for android .**


System.out.println("Diff" + duration + diffInDays + diffInHours + diffInMinutes + diffInSeconds); **// Print console message for Java .**

Java.time 持续时间

我仍然觉得没有一个答案是最新的和切中要点的。所以这里有一个现代的答案,使用 Java.time 中的 Duration,现代的 Java 日期和时间 API (MayurB 和 mkobit 的答案提到了同一个类,但没有一个正确地转换为天、小时、分钟和分钟)。

    DateTimeFormatter formatter = DateTimeFormatter.ofPattern("yy/MM/dd HH:mm:ss");
    

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


ZoneId zone = ZoneId.systemDefault();
ZonedDateTime startDateTime = LocalDateTime.parse(dateStart, formatter).atZone(zone);
ZonedDateTime endDateTime = LocalDateTime.parse(dateStop, formatter).atZone(zone);
    

Duration diff = Duration.between(startDateTime, endDateTime);
if (diff.isZero()) {
System.out.println("0 minutes");
} else {
long days = diff.toDays();
if (days != 0) {
System.out.print("" + days + " days ");
diff = diff.minusDays(days);
}
long hours = diff.toHours();
if (hours != 0) {
System.out.print("" + hours + " hours ");
diff = diff.minusHours(hours);
}
long minutes = diff.toMinutes();
if (minutes != 0) {
System.out.print("" + minutes + " minutes ");
diff = diff.minusMinutes(minutes);
}
long seconds = diff.getSeconds();
if (seconds != 0) {
System.out.print("" + seconds + " seconds ");
}
System.out.println();
}

Output from this example snippet is:

3分45秒

请注意,Duration总是将一天计算为24小时。如果你想用不同的方法处理时间异常,比如夏季时间过渡,解决方案包括: (1)使用 ChronoUnit.DAYS(2)使用 Period(3) Use LocalDateTimeinstead ofZonedDateTime’(可能被认为是一种技巧)。

上面的代码用于 Java8和 ThreeTen Backport,后者是 Java.time 到 Java6和7的 Backport。从 Java9开始,可以使用在那里添加的 toHoursParttoMinutesParttoSecondsPart方法更好地编写它。

等哪天我有时间的时候,我会进一步详细解释的,也许下周再说吧。

您可以使用以下命令获得两个 DateTime 之间的差异

DateTime startDate = DateTime.now();
DateTime endDate = DateTime.now();
Days daysBetween = Days.daysBetween(startDate, endDate);
System.out.println(daysBetween.toStandardSeconds());

下面的代码将给出两个 DateTime 之间的区别(将在 Java8及以上版本中工作)

private long countDaysBetween(LocalDateTime startDate, LocalDateTime enddate)
{
if(startDate == null || enddate == null)
{
throw new IllegalArgumentException("No such a date");
}
    

long daysBetween = ChronoUnit.DAYS.between(startDate, enddate);
    

return daysBetween;
}

如果有人想要一个包含所有字符串的字符串,可以使用这个函数。

String getTimeDifference(long duration) {
StringBuilder timeRemaining = new StringBuilder();


long days = TimeUnit.MILLISECONDS.toDays(duration);
if (days >= 1) {
timeRemaining.append(days).append((days == 1) ? " day " : " days ");
}


duration -= TimeUnit.DAYS.toMillis(days);
long hours = TimeUnit.MILLISECONDS.toHours(duration);
if (hours >= 1) {
timeRemaining.append(hours).append((hours == 1) ? " hour " : " hours ");
}


duration -= TimeUnit.HOURS.toMillis(hours);
long minutes = TimeUnit.MILLISECONDS.toMinutes(duration);
if (minutes >= 1) {
timeRemaining.append(minutes).append((hours == 1) ? " minute " : " minutes ");
}
        

return timeRemaining.toString().trim();
}