118

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，但不确定如何继续获得预期结果，如下所示：

示例：

      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

代码：

    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.");
4

18 回答 18

214

使用 Java 内置类TimeUnit可以更好地处理日期差异转换。它提供了实用方法来做到这一点：

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);
于 2013-07-30T07:19:39.070 回答
72

尝试以下

{
        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;
}
于 2013-11-10T06:58:45.007 回答
48

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());

另一个示例 日期差异

再来一个链接

或使用 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();
于 2013-07-30T06:58:17.617 回答
13

就像 shamimz 的回答一样，这是 Java 8 中如何解决问题的方法。

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)");

该代码产生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、秒的差异。

于 2014-06-04T22:19:39.877 回答
8

您可以创建一个方法，如

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。

于 2015-10-27T12:26:40.910 回答
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");
于 2013-11-19T03:40:22.073 回答
6

正如 Michael Borgwardt 在他的回答中所写：

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

请注意，这适用于 UTC 日期，因此如果您查看当地日期，差异可能是一天。由于夏令时，让它与当地日期正常工作需要完全不同的方法。

于 2013-07-30T06:58:30.890 回答
3

它对我有用，可以试试这个，希望它会有所帮助。让我知道是否有任何问题。

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 .**
于 2019-02-04T07:49:25.957 回答
3

在 Java 8 中，您可以制作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));
于 2015-11-17T20:30:31.697 回答
2

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);

出局是：

duration: 1 day 11 hour 15 min 9 sec
于 2015-11-19T11:53:02.387 回答
2

参考 shamim's answer update here 是一种无需使用任何第三方库即可完成任务的方法。只需复制方法并使用

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 :)
    }
于 2019-01-09T12:22:05.890 回答
2

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，它会打印：7 days ago.

于 2017-02-25T07:43:17.430 回答
1

java.time.Duration

我仍然觉得没有任何答案是最新的和中肯的。所以这里是使用Durationjava.time 的现代答案，现代 Java 日期和时间 API（MayurB 和 mkobit 的答案提到了同一个类，但它们都没有按照要求正确转换为天、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();
    }

此示例代码段的输出为：

3分45秒

请注意，Duration始终将一天视为 24 小时。如果您想以不同的方式处理诸如夏令时转换之类的时间异常，解决方案包括 (1) 使用ChronoUnit.DAYS(2) 使用Period(3) Use LocalDateTime instead ofZonedDateTime`（可能被视为黑客攻击）。

上面的代码适用于 Java 8 和 ThreeTen Backport，即 java.time 到 Java 6 和 7 的反向移植。从 Java 9 开始，可以使用方法更好地编写它toHoursPart，toMinutesPart并toSecondsPart添加到那里。

等我有时间的时候，我会进一步详细解释解释，也许要到下周。

于 2020-08-19T06:47:51.590 回答
0

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

于 2016-08-11T17:15:04.117 回答
0

如果有人想将所有字符串放在一起，可以使用此功能。

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();
}
于 2021-11-28T13:04:06.547 回答
0

您可以使用这个来获得两个 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());
于 2020-11-24T07:58:02.993 回答
0

下面的代码将给出两个 DateTime 之间的区别（将在 Java 8 及更高版本中工作）

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;
}
于 2021-04-28T11:58:50.800 回答
-2 
   // 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

于 2017-01-19T14:44:01.753 回答