如何在java中找到两个日期之间的差异持续时间?

我有两个DateTime对象,需要找到它们的差异持续时间,

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:

      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

    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.");
J888 asked 2019-08-13T13:02:31Z
14个解决方案
162 votes

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);
long diffInDays = TimeUnit.MILLISECONDS.toDays(duration);
Shamim Ahmmed answered 2019-08-13T13:03:08Z
63 votes

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;
}
answered 2019-08-13T13:02:44Z
44 votes

使用Joda-Time库

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

Joda Time有一个时间间隔的概念:

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

还有一个例子日期差异

还有一个链接

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();
MayurB answered 2019-08-13T13:03:56Z
12 votes

以下是Java 8中问题的解决方法,就像shamimz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)");

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)

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。

johnkarka answered 2019-08-13T13:04:47Z
6 votes

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());
}

此方法将返回2天之间的天数。

Vishal Kumar answered 2019-08-13T13:05:13Z
6 votes
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");
mkobit answered 2019-08-13T13:05:30Z
5 votes

正如Michael Borgwardt在他的回答中写道:

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

请注意,这适用于UTC日期,因此差异可能是一天   如果你看看当地的日期。 让它正常工作   由于日光,当地日期需要完全不同的方法   节省时间。

Ghostman answered 2019-08-13T13:06:09Z
3 votes

在Java 8中,您可以制作LocalDateTime,LocalDateTimeLocalDateTime.以下是一个示例:

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));
mkobit answered 2019-08-13T13:06:34Z
1 votes

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);

而外面是:

duration: 1 day 11 hour 15 min 9 sec
AYR1112013 answered 2019-08-13T13:07:01Z
1 votes

参考shamim的答案更新这里是一种在不使用任何第三方库的情况下完成任务的方法。 只需复制方法并使用即可

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 :)
    }
Tarun Kumar answered 2019-08-13T13:07:27Z
1 votes

它对我有用,可以试试这个,希望它会有所帮助。 如果有任何疑虑,请告诉我。

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 .**
Tarit Ray answered 2019-08-13T13:07:52Z
0 votes

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]

Naman Chhaparia answered 2019-08-13T13:08:26Z
0 votes

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();
    }

正如预期的那样它打印:7 days ago

i_am_zero answered 2019-08-13T13:09:02Z
-1 votes
   // 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);

如何计算Java中两个日期之间的差异

Sugriv Shekhar answered 2019-08-13T13:09:28Z
translate from https://stackoverflow.com:/questions/17940200/how-to-find-the-duration-of-difference-between-two-dates-in-java