Android/Java - 以天为单位的日期差异

Question

我正在使用以下代码获取当前日期（格式为 12/31/1999，即 mm/dd/yyyy）：

Textview txtViewData;
txtViewDate.setText("Today is " +
        android.text.format.DateFormat.getDateFormat(this).format(new Date()));

我还有另一个日期格式为：2010-08-25 (i.e. yyyy/mm/dd)，

所以我想找到日期与天数之间的差异，我如何找到天数之间的差异？

（换句话说，我想找出CURRENT DATE - yyyy/mm/dd 格式化日期之间的区别）

Answer 1

不是一个真正可靠的方法，最好使用JodaTime

  Calendar thatDay = Calendar.getInstance();
  thatDay.set(Calendar.DAY_OF_MONTH,25);
  thatDay.set(Calendar.MONTH,7); // 0-11 so 1 less
  thatDay.set(Calendar.YEAR, 1985);

  Calendar today = Calendar.getInstance();

  long diff = today.getTimeInMillis() - thatDay.getTimeInMillis(); //result in millis

这是一个近似值...

long days = diff / (24 * 60 * 60 * 1000);

要从字符串中解析日期，您可以使用

  String strThatDay = "1985/08/25";
  SimpleDateFormat formatter = new SimpleDateFormat("yyyy/MM/dd");
  Date d = null;
  try {
   d = formatter.parse(strThatDay);//catch exception
  } catch (ParseException e) {
   // TODO Auto-generated catch block
   e.printStackTrace();
  } 


  Calendar thatDay = Calendar.getInstance();
  thatDay.setTime(d); //rest is the same....

不过，既然您确定日期格式... 您也可以对它的子字符串执行Integer.parseInt() 来获取它们的数值。

Answer 2

这不是我的工作，找到了答案here。不希望将来出现断开的链接:)。

关键是考虑到日光设置的这一行，参考完整代码。

TimeZone.setDefault(TimeZone.getTimeZone("Europe/London"));

或尝试将TimeZone 作为参数传递给daysBetween()，并在sDate 和eDate 对象中调用setTimeZone()。

就这样吧：

public static Calendar getDatePart(Date date){
    Calendar cal = Calendar.getInstance();       // get calendar instance
    cal.setTime(date);      
    cal.set(Calendar.HOUR_OF_DAY, 0);            // set hour to midnight
    cal.set(Calendar.MINUTE, 0);                 // set minute in hour
    cal.set(Calendar.SECOND, 0);                 // set second in minute
    cal.set(Calendar.MILLISECOND, 0);            // set millisecond in second
    
    return cal;                                  // return the date part
}

getDatePart() 取自here

/**
 * This method also assumes endDate >= startDate
**/
public static long daysBetween(Date startDate, Date endDate) {
  Calendar sDate = getDatePart(startDate);
  Calendar eDate = getDatePart(endDate);

  long daysBetween = 0;
  while (sDate.before(eDate)) {
      sDate.add(Calendar.DAY_OF_MONTH, 1);
      daysBetween++;
  }
  return daysBetween;
}

细微差别： 找出两个日期之间的差异并不像减去两个日期并将结果除以 (24 * 60 * 60 * 1000) 那样简单。事实上，它是错误的！

例如： 03/24/2007 和 03/25/2007 这两个日期之间的差应该是 1 天；但是，使用上述方法，在英国，您将获得 0 天！

自己看看（代码如下）。以毫秒为单位会导致四舍五入的错误，一旦你有像夏令时这样的小东西，它们就会变得最明显。

完整代码：

import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.Date;
import java.util.TimeZone;

public class DateTest {

public class DateTest {

static SimpleDateFormat sdf = new SimpleDateFormat("dd-MMM-yyyy");

public static void main(String[] args) {

  TimeZone.setDefault(TimeZone.getTimeZone("Europe/London"));

  //diff between these 2 dates should be 1
  Date d1 = new Date("01/01/2007 12:00:00");
  Date d2 = new Date("01/02/2007 12:00:00");

  //diff between these 2 dates should be 1
  Date d3 = new Date("03/24/2007 12:00:00");
  Date d4 = new Date("03/25/2007 12:00:00");

  Calendar cal1 = Calendar.getInstance();cal1.setTime(d1);
  Calendar cal2 = Calendar.getInstance();cal2.setTime(d2);
  Calendar cal3 = Calendar.getInstance();cal3.setTime(d3);
  Calendar cal4 = Calendar.getInstance();cal4.setTime(d4);

  printOutput("Manual   ", d1, d2, calculateDays(d1, d2));
  printOutput("Calendar ", d1, d2, daysBetween(cal1, cal2));
  System.out.println("---");
  printOutput("Manual   ", d3, d4, calculateDays(d3, d4));
  printOutput("Calendar ", d3, d4, daysBetween(cal3, cal4));
}


private static void printOutput(String type, Date d1, Date d2, long result) {
  System.out.println(type+ "- Days between: " + sdf.format(d1)
                    + " and " + sdf.format(d2) + " is: " + result);
}

/** Manual Method - YIELDS INCORRECT RESULTS - DO NOT USE**/
/* This method is used to find the no of days between the given dates */
public static long calculateDays(Date dateEarly, Date dateLater) {
  return (dateLater.getTime() - dateEarly.getTime()) / (24 * 60 * 60 * 1000);
}

/** Using Calendar - THE CORRECT WAY**/
public static long daysBetween(Date startDate, Date endDate) {
  ...
}

输出：

手动 - 2007 年 1 月 1 日至 2007 年 1 月 2 日之间的天数为：1

日历 - 2007 年 1 月 1 日至 2007 年 1 月 2 日之间的天数为：1

---

手动 - 2007 年 3 月 24 日和 2007 年 3 月 25 日之间的天数为：0

日历 - 2007 年 3 月 24 日和 2007 年 3 月 25 日之间的天数为：1

Answer 3

大多数答案都很好，适合您的问题

所以我想找出日期与天数之间的差异，我如何找到天数之间的差异？

我建议采用这种非常简单直接的方法，保证在任何时区都能为您提供正确的差异：

int difference= 
((int)((startDate.getTime()/(24*60*60*1000))
-(int)(endDate.getTime()/(24*60*60*1000))));

就是这样！

Answer 4

使用jodatime API

Days.daysBetween(start.toDateMidnight() , end.toDateMidnight() ).getDays() 

'start' 和 'end' 是你的 DateTime 对象。要将您的日期字符串解析为 DateTime 对象，请使用 parseDateTime method

还有一个android specific JodaTime library。

Answer 5

这个片段考虑了夏令时并且是 O(1)。

private final static long MILLISECS_PER_DAY = 24 * 60 * 60 * 1000;

private static long getDateToLong(Date date) {
    return Date.UTC(date.getYear(), date.getMonth(), date.getDate(), 0, 0, 0);
}

public static int getSignedDiffInDays(Date beginDate, Date endDate) {
    long beginMS = getDateToLong(beginDate);
    long endMS = getDateToLong(endDate);
    long diff = (endMS - beginMS) / (MILLISECS_PER_DAY);
    return (int)diff;
}

public static int getUnsignedDiffInDays(Date beginDate, Date endDate) {
    return Math.abs(getSignedDiffInDays(beginDate, endDate));
}

Answer 6

这对我来说是简单且最佳的计算，可能也适合你。

       try {
            /// String CurrDate=  "10/6/2013";
            /// String PrvvDate=  "10/7/2013";
            Date date1 = null;
            Date date2 = null;
            SimpleDateFormat df = new SimpleDateFormat("M/dd/yyyy");
            date1 = df.parse(CurrDate);
            date2 = df.parse(PrvvDate);
            long diff = Math.abs(date1.getTime() - date2.getTime());
            long diffDays = diff / (24 * 60 * 60 * 1000);


            System.out.println(diffDays);

        } catch (Exception e1) {
            System.out.println("exception " + e1);
        }

Answer 7

tl;博士

ChronoUnit.DAYS.between( 
    LocalDate.parse( "1999-12-28" ) , 
    LocalDate.parse( "12/31/1999" , DateTimeFormatter.ofPattern( "MM/dd/yyyy" ) ) 
)

详情

其他答案已过时。与 Java 的最早版本捆绑在一起的旧日期时间类已被证明设计不佳、混乱且麻烦。避开他们。

java.time

Joda-Time 项目作为旧课程的替代品非常成功。这些类为 Java 8 及更高版本中内置的java.time 框架提供了灵感。

大部分 java.time 功能在ThreeTen-Backport 中向后移植到 Java 6 和 7，并在 ThreeTenABP 中进一步适应 Android。

LocalDate

LocalDate 类表示没有时间和时区的仅日期值。

解析字符串

如果您输入的字符串是标准的ISO 8601 格式，LocalDate 类可以直接解析字符串。

LocalDate start = LocalDate.parse( "1999-12-28" );

如果不是 ISO 8601 格式，请使用 DateTimeFormatter 定义格式模式。

String input = "12/31/1999";
DateTimeFormatter formatter = DateTimeFormatter.ofPattern( "MM/dd/yyyy" );
LocalDate stop = LocalDate.parse( input , formatter );

通过ChronoUnit 的经过天数

现在计算这对 LocalDate 对象之间经过的天数。 ChronoUnit 枚举计算经过的时间。

long totalDays = ChronoUnit.DAYS.between( start , stop ) ; 

如果您不熟悉 Java 枚举，请知道它们比大多数其他编程语言中的传统枚举更强大和有用。请参阅 Enum 类文档、Oracle Tutorial 和 Wikipedia 了解更多信息。

---

关于java.time

java.time 框架内置于 Java 8 及更高版本中。这些类取代了麻烦的旧 legacy 日期时间类，例如 java.util.Date、Calendar 和 SimpleDateFormat。

Joda-Time 项目现在位于maintenance mode，建议迁移到java.time 类。

要了解更多信息，请参阅Oracle Tutorial。并在 Stack Overflow 上搜索许多示例和解释。规格为JSR 310。

从哪里获得 java.time 类？

• Java SE 8 和 SE 9 及更高版本
  • 内置。
  • 标准 Java API 的一部分，带有捆绑实现。
  • Java 9 添加了一些小功能和修复。
• Java SE 6 和 SE 7
  • ThreeTen-Backport 中的大部分 java.time 功能都向后移植到 Java 6 和 7。
• Android
  • ThreeTenABP 项目专门针对 Android 改编了 ThreeTen-Backport（如上所述）。
  • 见How to use ThreeTenABP…。

ThreeTen-Extra 项目通过附加类扩展了 java.time。该项目是未来可能添加到 java.time 的试验场。您可以在这里找到一些有用的类，例如Interval、YearWeek、YearQuarter 和more。

Answer 8

Sam Quest 的答案中的 Correct Way 仅在第一个日期早于第二个日期时才有效。此外，如果两个日期在一天之内，它将返回 1。

这是最适合我的解决方案。就像大多数其他解决方案一样，由于夏令时偏移错误，它仍然会在一年中的两天显示不正确的结果。

private final static long MILLISECS_PER_DAY = 24 * 60 * 60 * 1000;

long calculateDeltaInDays(Calendar a, Calendar b) {

    // Optional: avoid cloning objects if it is the same day
    if(a.get(Calendar.ERA) == b.get(Calendar.ERA) 
            && a.get(Calendar.YEAR) == b.get(Calendar.YEAR)
            && a.get(Calendar.DAY_OF_YEAR) == b.get(Calendar.DAY_OF_YEAR)) {
        return 0;
    }
    Calendar a2 = (Calendar) a.clone();
    Calendar b2 = (Calendar) b.clone();
    a2.set(Calendar.HOUR_OF_DAY, 0);
    a2.set(Calendar.MINUTE, 0);
    a2.set(Calendar.SECOND, 0);
    a2.set(Calendar.MILLISECOND, 0);
    b2.set(Calendar.HOUR_OF_DAY, 0);
    b2.set(Calendar.MINUTE, 0);
    b2.set(Calendar.SECOND, 0);
    b2.set(Calendar.MILLISECOND, 0);
    long diff = a2.getTimeInMillis() - b2.getTimeInMillis();
    long days = diff / MILLISECS_PER_DAY;
    return Math.abs(days);
}

Answer 9

最好和最简单的方法

  public int getDays(String begin) throws ParseException {
     long MILLIS_PER_DAY = 24 * 60 * 60 * 1000;
     SimpleDateFormat dateFormat = new SimpleDateFormat("dd-MM-yyyy", Locale.ENGLISH);

    long begin = dateFormat.parse(begin).getTime();
    long end = new Date().getTime(); // 2nd date want to compare
    long diff = (end - begin) / (MILLIS_PER_DAY);
    return (int) diff;
}

Answer 10

使用以下函数：

   /**
     * Returns the number of days between two dates. The time part of the
     * days is ignored in this calculation, so 2007-01-01 13:00 and 2007-01-02 05:00
     * have one day inbetween.
     */
    public static long daysBetween(Date firstDate, Date secondDate) {
        // We only use the date part of the given dates
        long firstSeconds = truncateToDate(firstDate).getTime()/1000;
        long secondSeconds = truncateToDate(secondDate).getTime()/1000;
        // Just taking the difference of the millis.
        // These will not be exactly multiples of 24*60*60, since there
        // might be daylight saving time somewhere inbetween. However, we can
        // say that by adding a half day and rounding down afterwards, we always
        // get the full days.
        long difference = secondSeconds-firstSeconds;
        // Adding half a day
        if( difference >= 0 ) {
            difference += SECONDS_PER_DAY/2; // plus half a day in seconds
        } else {
            difference -= SECONDS_PER_DAY/2; // minus half a day in seconds
        }
        // Rounding down to days
        difference /= SECONDS_PER_DAY;

        return difference;
    }

    /**
     * Truncates a date to the date part alone.
     */
    @SuppressWarnings("deprecation")
    public static Date truncateToDate(Date d) {
        if( d instanceof java.sql.Date ) {
            return d; // java.sql.Date is already truncated to date. And raises an
                      // Exception if we try to set hours, minutes or seconds.
        }
        d = (Date)d.clone();
        d.setHours(0);
        d.setMinutes(0);
        d.setSeconds(0);
        d.setTime(((d.getTime()/1000)*1000));
        return d;
    }

Answer 11

有一个简单的解决方案，至少对我来说，是唯一可行的解​​决方案。

问题在于，我看到的所有答案——使用 Joda、Calendar、Date 或其他任何东西——只考虑了毫秒数。他们最终计算的是两个日期之间的 24 小时周期数，而不是实际的天数。因此，从 1 月 1 日晚上 11 点到 1 月 2 日凌晨 1 点的时间将返回 0 天。

要计算startDate 和endDate 之间的实际天数，只需执行以下操作：

// Find the sequential day from a date, essentially resetting time to start of the day
long startDay = startDate.getTime() / 1000 / 60 / 60 / 24;
long endDay = endDate.getTime() / 1000 / 60 / 60 / 24;

// Find the difference, duh
long daysBetween = endDay - startDay;

这将在 1 月 2 日和 1 月 1 日之间返回“1”。如果您需要计算结束日，只需将 1 加到 daysBetween （我需要在我的代码中这样做，因为我想计算范围内的总天数）。

这有点类似于Daniel has suggested，但我认为代码更小。

Answer 12

所有这些解决方案都存在以下两个问题之一。由于舍入误差、闰日和秒数等原因，解决方案并不完全准确，或者您最终会循环计算两个未知日期之间的天数。

此解决方案解决了第一个问题，并将第二个问题提高了大约 365 倍，如果您知道最大范围是多少，效果会更好。

/**
 * @param thisDate
 * @param thatDate
 * @param maxDays
 *            set to -1 to not set a max
 * @returns number of days covered between thisDate and thatDate, inclusive, i.e., counting both
 *          thisDate and thatDate as an entire day. Will short out if the number of days exceeds
 *          or meets maxDays
 */
public static int daysCoveredByDates(Date thisDate, Date thatDate, int maxDays) {
    //Check inputs
    if (thisDate == null || thatDate == null) {
        return -1;
    }

    //Set calendar objects
    Calendar startCal = Calendar.getInstance();
    Calendar endCal = Calendar.getInstance();
    if (thisDate.before(thatDate)) {
        startCal.setTime(thisDate);
        endCal.setTime(thatDate);
    }
    else {
        startCal.setTime(thatDate);
        endCal.setTime(thisDate);
    }

    //Get years and dates of our times.
    int startYear = startCal.get(Calendar.YEAR);
    int endYear = endCal.get(Calendar.YEAR);
    int startDay = startCal.get(Calendar.DAY_OF_YEAR);
    int endDay = endCal.get(Calendar.DAY_OF_YEAR);

    //Calculate the number of days between dates.  Add up each year going by until we catch up to endDate.
    while (startYear < endYear && maxDays >= 0 && endDay - startDay + 1 < maxDays) {
        endDay += startCal.getActualMaximum(Calendar.DAY_OF_YEAR); //adds the number of days in the year startDate is currently in
        ++startYear;
        startCal.set(Calendar.YEAR, startYear); //reup the year
    }
    int days = endDay - startDay + 1;

    //Honor the maximum, if set
    if (maxDays >= 0) {
        days = Math.min(days, maxDays);
    }
    return days;
}

如果您需要日期之间的天数（不包括后一个日期），请在看到 endDay - startDay + 1 时去掉 + 1。

Answer 13

另一种方式：

public static int numberOfDaysBetweenDates(Calendar fromDay, Calendar toDay) {
        fromDay = calendarStartOfDay(fromDay);
        toDay = calendarStartOfDay(toDay);
        long from = fromDay.getTimeInMillis();
        long to = toDay.getTimeInMillis();
        return (int) TimeUnit.MILLISECONDS.toDays(to - from);
    }

Answer 14

        Date userDob = new SimpleDateFormat("yyyy-MM-dd").parse(dob);
        Date today = new Date();
        long diff =  today.getTime() - userDob.getTime();
        int numOfDays = (int) (diff / (1000 * 60 * 60 * 24));
        int hours = (int) (diff / (1000 * 60 * 60));
        int minutes = (int) (diff / (1000 * 60));
        int seconds = (int) (diff / (1000));

Answer 15

使用这些功能

    public static int getDateDifference(int previousYear, int previousMonthOfYear, int previousDayOfMonth, int nextYear, int nextMonthOfYear, int nextDayOfMonth, int differenceToCount){
    // int differenceToCount = can be any of the following
    //  Calendar.MILLISECOND;
    //  Calendar.SECOND;
    //  Calendar.MINUTE;
    //  Calendar.HOUR;
    //  Calendar.DAY_OF_MONTH;
    //  Calendar.MONTH;
    //  Calendar.YEAR;
    //  Calendar.----

    Calendar previousDate = Calendar.getInstance();
    previousDate.set(Calendar.DAY_OF_MONTH, previousDayOfMonth);
    // month is zero indexed so month should be minus 1
    previousDate.set(Calendar.MONTH, previousMonthOfYear);
    previousDate.set(Calendar.YEAR, previousYear);

    Calendar nextDate = Calendar.getInstance();
    nextDate.set(Calendar.DAY_OF_MONTH, previousDayOfMonth);
    // month is zero indexed so month should be minus 1
    nextDate.set(Calendar.MONTH, previousMonthOfYear);
    nextDate.set(Calendar.YEAR, previousYear);

    return getDateDifference(previousDate,nextDate,differenceToCount);
}
public static int getDateDifference(Calendar previousDate,Calendar nextDate,int differenceToCount){
    // int differenceToCount = can be any of the following
    //  Calendar.MILLISECOND;
    //  Calendar.SECOND;
    //  Calendar.MINUTE;
    //  Calendar.HOUR;
    //  Calendar.DAY_OF_MONTH;
    //  Calendar.MONTH;
    //  Calendar.YEAR;
    //  Calendar.----

    //raise an exception if previous is greater than nextdate.
    if(previousDate.compareTo(nextDate)>0){
        throw new RuntimeException("Previous Date is later than Nextdate");
    }

    int difference=0;
    while(previousDate.compareTo(nextDate)<=0){
        difference++;
        previousDate.add(differenceToCount,1);
    }
    return difference;
}

Answer 16

        public void dateDifferenceExample() {

        // Set the date for both of the calendar instance
        GregorianCalendar calDate = new GregorianCalendar(2012, 10, 02,5,23,43);
        GregorianCalendar cal2 = new GregorianCalendar(2015, 04, 02);

        // Get the represented date in milliseconds
        long millis1 = calDate.getTimeInMillis();
        long millis2 = cal2.getTimeInMillis();

        // Calculate difference in milliseconds
        long diff = millis2 - millis1;

        // Calculate difference in seconds
        long diffSeconds = diff / 1000;

        // Calculate difference in minutes
        long diffMinutes = diff / (60 * 1000);

        // Calculate difference in hours
        long diffHours = diff / (60 * 60 * 1000);

        // Calculate difference in days
        long diffDays = diff / (24 * 60 * 60 * 1000);
    Toast.makeText(getContext(), ""+diffSeconds, Toast.LENGTH_SHORT).show();


}

Answer 17

我找到了一种非常简单的方法来做到这一点，这就是我在我的应用程序中使用的方法。

假设您在 Time 对象中有日期（或者其他什么，我们只需要毫秒）：

Time date1 = initializeDate1(); //get the date from somewhere
Time date2 = initializeDate2(); //get the date from somewhere

long millis1 = date1.toMillis(true);
long millis2 = date2.toMillis(true);

long difference = millis2 - millis1 ;

//now get the days from the difference and that's it
long days = TimeUnit.MILLISECONDS.toDays(difference);

//now you can do something like
if(days == 7)
{
    //do whatever when there's a week of difference
}

if(days >= 30)
{
    //do whatever when it's been a month or more
}

Answer 18

乔达时间

最好的方法是使用Joda-Time，这是您可以添加到项目中的非常成功的开源库。

String date1 = "2015-11-11";
String date2 = "2013-11-11";
DateTimeFormatter formatter = new DateTimeFormat.forPattern("yyyy-MM-dd");
DateTime d1 = formatter.parseDateTime(date1);
DateTime d2 = formatter.parseDateTime(date2);
long diffInMillis = d2.getMillis() - d1.getMillis();

Duration duration = new Duration(d1, d2);
int days = duration.getStandardDays();
int hours = duration.getStandardHours();
int minutes = duration.getStandardMinutes();

如果你使用Android Studio，很容易添加joda-time。在您的 build.gradle（应用程序）中：

dependencies {
  compile 'joda-time:joda-time:2.4'
  compile 'joda-time:joda-time:2.4'
  compile 'joda-time:joda-time:2.2'
}