如何在java中找到两个日期之间的差异持续时间?

时间:2021-06-18 21:28:23

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

我有两个DateTime对象,需要找到它们的差异持续时间,

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

我有以下代码,但不知道如何继续它以获得预期的结果如下:

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 mintues

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.");

12 个解决方案

#1


57  

try the following

尝试以下方法

{
        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;
}

#2


129  

The date difference conversion could be handled in a better way using Java built-in class, TimeUnit. It provides utility methods to do that:

可以使用Java内置类TimeUnit以更好的方式处理日期差异转换。它提供了实用方法:

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);

#3


38  

Use Joda-Time library

使用Joda-Time库

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

Joda Time has a concept of time Interval:

Joda Time有一个时间间隔的概念:

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

One more example Date Difference

另一个例子日期差异

One more Link

还有一个链接

or with Java-8 (which integrated Joda-Time concepts)

或者使用Java-8(集成了Joda-Time概念)

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

#4


10  

Here is how the problem can solved in Java 8 just like the answer by shamimz.

以下是Java 8中问题如何解决,就像shamimz的答案一样。

Source : http://docs.oracle.com/javase/tutorial/datetime/iso/period.html

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)");

The code produces output similar to the following:

代码生成类似于以下内容的输出:

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

We have to use LocalDateTime http://docs.oracle.com/javase/8/docs/api/java/time/LocalDateTime.html to get hour,minute,second differences.

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

#5


7  

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");

#6


6  

As Michael Borgwardt writes in his answer here:

正如Michael Borgwardt在他的回答中写道:

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 off if you look at local dates. And getting it to work correctly with local dates requires a completely different approach due to daylight savings time.

请注意,这适用于UTC日期,因此如果您查看本地日期,差异可能是休息日。由于夏令时,使其与本地日期一起正常工作需要完全不同的方法。

#7


6  

You can create a method like

你可以创建一个类似的方法

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

This method will return the number of days between the 2 days.

此方法将返回2天之间的天数。

#8


2  

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

在Java 8中,您可以创建DateTimeFormatter,Duration和LocalDateTime。这是一个例子:

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));

#9


1  

This is the code:

这是代码:

        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);

and the out is:

而外面是:

duration: 1 day 11 hour 15 min 9 sec

#10


0  

This is a program I wrote, which gets the number of days between 2 dates(no time here).

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

http://pastebin.com/HRsjTtUf

#11


0  

I solved the similar problem using a simple method recently.

我最近使用一种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();
    }

And as expected it prints: 7 days ago.

正如预期的那样它打印:7天前。

#12


-1  

   // 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);

How to calculate difference between two dates in Java

如何计算Java中两个日期之间的差异

#1


57  

try the following

尝试以下方法

{
        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;
}

#2


129  

The date difference conversion could be handled in a better way using Java built-in class, TimeUnit. It provides utility methods to do that:

可以使用Java内置类TimeUnit以更好的方式处理日期差异转换。它提供了实用方法:

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);

#3


38  

Use Joda-Time library

使用Joda-Time库

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

Joda Time has a concept of time Interval:

Joda Time有一个时间间隔的概念:

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

One more example Date Difference

另一个例子日期差异

One more Link

还有一个链接

or with Java-8 (which integrated Joda-Time concepts)

或者使用Java-8(集成了Joda-Time概念)

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

#4


10  

Here is how the problem can solved in Java 8 just like the answer by shamimz.

以下是Java 8中问题如何解决,就像shamimz的答案一样。

Source : http://docs.oracle.com/javase/tutorial/datetime/iso/period.html

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)");

The code produces output similar to the following:

代码生成类似于以下内容的输出:

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

We have to use LocalDateTime http://docs.oracle.com/javase/8/docs/api/java/time/LocalDateTime.html to get hour,minute,second differences.

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

#5


7  

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");

#6


6  

As Michael Borgwardt writes in his answer here:

正如Michael Borgwardt在他的回答中写道:

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 off if you look at local dates. And getting it to work correctly with local dates requires a completely different approach due to daylight savings time.

请注意,这适用于UTC日期,因此如果您查看本地日期,差异可能是休息日。由于夏令时,使其与本地日期一起正常工作需要完全不同的方法。

#7


6  

You can create a method like

你可以创建一个类似的方法

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

This method will return the number of days between the 2 days.

此方法将返回2天之间的天数。

#8


2  

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

在Java 8中,您可以创建DateTimeFormatter,Duration和LocalDateTime。这是一个例子:

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));

#9


1  

This is the code:

这是代码:

        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);

and the out is:

而外面是:

duration: 1 day 11 hour 15 min 9 sec

#10


0  

This is a program I wrote, which gets the number of days between 2 dates(no time here).

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

http://pastebin.com/HRsjTtUf

#11


0  

I solved the similar problem using a simple method recently.

我最近使用一种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();
    }

And as expected it prints: 7 days ago.

正如预期的那样它打印:7天前。

#12


-1  

   // 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);

How to calculate difference between two dates in Java

如何计算Java中两个日期之间的差异