65

我在这里看到了所有的帖子,但我仍然无法弄清楚两个安卓日期之间的差异。

这就是我所做的:

long diff = date1.getTime() - date2.getTime();
Date diffDate = new Date(diff);

我得到:日期是 1970 年 1 月 1 日,两个小时内的时间总是更长……我来自以色列,所以两个小时是 timeOffset。

我怎样才能得到正常的差异???

4

14 回答 14

143

你接近正确的答案,你得到了这两个日期之间的毫秒差异,但是当你试图从这个差异中构造一个日期时,它假设你想要创建一个新的Date对象,该差异值作为它的纪元时间。如果您正在寻找以小时为单位的时间,那么您只需要对其进行一些基本的算术运算diff即可获得不同的时间部分。

爪哇:

long diff = date1.getTime() - date2.getTime();
long seconds = diff / 1000;
long minutes = seconds / 60;
long hours = minutes / 60;
long days = hours / 24;

科特林:

val diff: Long = date1.getTime() - date2.getTime()
val seconds = diff / 1000
val minutes = seconds / 60
val hours = minutes / 60
val days = hours / 24

所有这些数学运算都将简单地进行整数运算,因此它将截断任何小数点

于 2012-05-21T18:24:16.067 回答
56
    long diffInMillisec = date1.getTime() - date2.getTime();

    long diffInDays = TimeUnit.MILLISECONDS.toDays(diffInMillisec);
    long diffInHours = TimeUnit.MILLISECONDS.toHours(diffInMillisec);
    long diffInMin = TimeUnit.MILLISECONDS.toMinutes(diffInMillisec);
    long diffInSec = TimeUnit.MILLISECONDS.toSeconds(diffInMillisec);
于 2016-02-09T13:51:19.703 回答
23

一些补充: 在这里我将字符串转换为日期,然后比较当前时间。

String toyBornTime = "2014-06-18 12:56:50";
    SimpleDateFormat dateFormat = new SimpleDateFormat(
            "yyyy-MM-dd HH:mm:ss");

    try {

        Date oldDate = dateFormat.parse(toyBornTime);
        System.out.println(oldDate);

        Date currentDate = new Date();

        long diff = currentDate.getTime() - oldDate.getTime();
        long seconds = diff / 1000;
        long minutes = seconds / 60;
        long hours = minutes / 60;
        long days = hours / 24;

        if (oldDate.before(currentDate)) {

            Log.e("oldDate", "is previous date");
            Log.e("Difference: ", " seconds: " + seconds + " minutes: " + minutes
                    + " hours: " + hours + " days: " + days);

        }

        // Log.e("toyBornTime", "" + toyBornTime);

    } catch (ParseException e) {

        e.printStackTrace();
    }
于 2014-06-18T07:10:09.973 回答
10

java.time.Duration

使用java.time.Duration

    Duration diff = Duration.between(instant2, instant1);
    System.out.println(diff);

这将打印类似

PT109H27M21S

这意味着 109 小时 27 分 21 秒的时间段。如果你想要一些更易读的东西——我会先给出 Java 9 版本,它最简单:

    String formattedDiff = String.format(Locale.ENGLISH,
            "%d days %d hours %d minutes %d seconds",
            diff.toDays(), diff.toHoursPart(), diff.toMinutesPart(), diff.toSecondsPart());
    System.out.println(formattedDiff);

现在我们得到

4 days 13 hours 27 minutes 21 seconds

该类Durationjava.time现代 Java 日期和时间 API 的一部分。这捆绑在较新的 Android 设备上。在旧设备上,获取 ThreeTenABP 并将其添加到您的项目中,并确保org.threeten.bp.Duration从同一个包中导入您可能需要的其他日期时间类。

假设您还没有获得 Java 9 版本,您可以依次减去较大的单位以获得较小的单位:

    long days = diff.toDays();
    diff = diff.minusDays(days);
    long hours = diff.toHours();
    diff = diff.minusHours(hours);
    long minutes = diff.toMinutes();
    diff = diff.minusMinutes(minutes);
    long seconds = diff.toSeconds();

然后你可以像上面那样格式化四个变量。

你做错了什么?

ADate代表一个时间点。它从来都不是用来代表时间量、持续时间的,也不适合它。试图使这项工作充其量只会导致混乱和难以维护的代码。你不想这样,所以请不要。

问题:java.time 不需要 Android API 级别 26 吗?

java.time 在较旧和较新的 Android 设备上都能很好地工作。它只需要至少Java 6

  • 在 Java 8 及更高版本以及更新的 Android 设备(从 API 级别 26 开始)中,现代 API 是内置的。
  • 在非 Android Java 6 和 7 中获得 ThreeTen Backport,现代类的后向端口(ThreeTen 用于 JSR 310;请参阅底部的链接)。
  • 在(较旧的)Android 上使用 ThreeTen Backport 的 Android 版本。它被称为 ThreeTenABP。并确保从org.threeten.bp子包中导入日期和时间类。

链接

于 2018-01-30T20:28:12.487 回答
4

使用Instant怎么样:

val instant1 = now()
val instant2 = now()
val diff: Duration = Duration.between(instant1, instant2)
val minutes = diff.toMinutes()

您甚至可以使用 Instant1.toString() 保存 Instant 并使用 parse(string) 解析该字符串。

如果您需要支持 Android API 级别 < 26,只需 在您的项目中添加Java 8+ API 脱糖支持。

于 2021-04-12T12:44:23.490 回答
3

如果你使用 Kotlin 语言进行 Android 开发,你可以在这样的日子里获得不同:

fun daysDiff(c1: Calendar, c2: Calendar): Long {
    val diffInMillis = c1.timeInMillis - c2.timeInMillis
    return diffInMillis.milliseconds.inWholeDays
}

或者,如果您想获得不同类型的结果,您可以替换inWholeDaystoInt(...), toDouble(...), toString(...)

如果您感兴趣的唯一日期差异是天数(假设是 Double 类型的结果),您可以像这样创建运算符扩展方法:

operator fun Calendar.minus(c: Calendar): Double {
    val diffInMillis = timeInMillis - c.timeInMillis
    return diffInMillis.milliseconds.toDouble(DurationUnit.DAYS)
}

然后任何类似的操作calendar2 - calendar1都会以双倍的天数返回差异。

于 2020-05-11T13:24:07.870 回答
2

这是我基于@Ole VV 回答的回答。

这也适用于单数。

private String getDuration(Date d1, Date d2) {
    Duration diff = Duration.between(d1.toInstant(), d2.toInstant());


    long days = diff.toDays();
    diff = diff.minusDays(days);
    long hours = diff.toHours();
    diff = diff.minusHours(hours);
    long minutes = diff.toMinutes();
    diff = diff.minusMinutes(minutes);
    long seconds = diff.toMillis();

    StringBuilder formattedDiff = new StringBuilder();
    if(days!=0){
        if(days==1){
            formattedDiff.append(days + " Day ");

        }else {
            formattedDiff.append(days + " Days ");
        }
    }if(hours!=0){
        if(hours==1){
            formattedDiff.append(hours + " hour ");
        }else{
            formattedDiff.append(hours + " hours ");
        }
    }if(minutes!=0){
        if(minutes==1){
            formattedDiff.append(minutes + " minute ");
        }else{
            formattedDiff.append(minutes + " minutes ");
        }
    }if(seconds!=0){
        if(seconds==1){
            formattedDiff.append(seconds + " second ");
        }else{
            formattedDiff.append(seconds + " seconds ");
        }
    }


    return formattedDiff.toString();
}

它与 StringBuilder 一起工作以将所有内容附加在一起。

于 2019-07-18T09:02:27.710 回答
2

我试过这个方法..但不知道为什么我没有得到正确的结果

long diff = date1.getTime() - date2.getTime();
long seconds = diff / 1000;
long minutes = seconds / 60;
long hours = minutes / 60;
long days = hours / 24;

但这有效

long miliSeconds = date1.getTime() -date2.getTime();
long seconds = TimeUnit.MILLISECONDS.toSeconds(miliSeconds);
long minute = seconds/60;
long hour = minute/60;
long days = hour/24;
于 2020-08-01T16:30:11.250 回答
1

使用格鲁吉亚压延机

 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();


}
于 2019-01-07T07:35:19.687 回答
0

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;
}
于 2017-11-29T06:16:32.750 回答
0

用 Kotlin 编写: 如果您需要两个日期之间的差异并且不关心日期本身(如果您需要在应用程序中做某事,例如基于保存在共享首选项中的其他操作时间的时间)。第一次保存:

val firstTime:Long= System.currentTimeMillis()

第二次保存:

val now:Long= System.currentTimeMillis()

计算 2 次之间的毫秒数:

val milisecondsSinceLastTime: Long =(now-lastScrollTime)
于 2020-02-02T01:14:13.253 回答
0

获取上午和下午之间的时差

private int  getTimeDiff() {
    SimpleDateFormat sdf = new SimpleDateFormat("hh:mm:ss aa");
    Date systemDate = Calendar.getInstance().getTime();
    String myDate = sdf.format(systemDate);
    Date current = null;
    Date Date2 = null;
    try {
        current = sdf.parse(myDate);
        // current = sdf.parse("05:00:00 pm");
        Date2 = sdf.parse("12:00:00 am");
    } catch (ParseException e) {
        e.printStackTrace();
    }
    long millse = Date2.getTime() - current.getTime();
    long mills = Math.abs(millse);
    int Hours = (int) (mills / (1000 * 60 * 60));
    int Mins = (int) (mills / (1000 * 60)) % 60;
    long Secs = (int) (mills / 1000) % 60;
    String diff = Hours + ":" + Mins + ":" + Secs;
    int min = 60-Mins;
    if (Hours >= 12) {
        String requiredTime = 24 - Hours + ":" + Mins + ":" + Secs;
        int minutes= ((24-Hours)*60)-Mins;
        return minutes;

    } else {
        int time = 12 - Hours;
        int hours=time+12;
        int res = (hours*60);
        int minutes = res-Mins;
        return minutes;
    }
}
于 2022-02-05T18:10:41.923 回答
0

当您尝试根据该差异构造日期时,假设您要创建一个新的 Date 对象,该差异值作为其纪元时间。

//get time in milliseconds
long diff = date1.getTime() - date2.getTime();
//get time in seconds
long seconds = diff / 1000;
//and so on
long minutes = seconds / 60;
long hours = minutes / 60;
long days = hours / 24;

有关更多信息,您可以使用以下链接:

Oracle 教程:日期时间,解释如何使用 java.time。

ThreeTen Backport 项目,java.time 到 Java 6 和 7 的反向移植。

ThreeTenABP , ThreeTen Backport 的 Android 版本。

问题:如何在Android项目中使用ThreeTenABP,有非常详尽的解释。

于 2022-02-05T20:09:36.273 回答
-1

对我有用的最短答案。以毫秒为单位发送开始和结束日期。

public int GetDifference(long start,long end){
    Calendar cal = Calendar.getInstance();
    cal.setTimeInMillis(start);
    int hour = cal.get(Calendar.HOUR_OF_DAY);
    int min = cal.get(Calendar.MINUTE);
    long t=(23-hour)*3600000+(59-min)*60000;

    t=start+t;

    int diff=0;
    if(end>t){
        diff=(int)((end-t)/ TimeUnit.DAYS.toMillis(1))+1;
    }

    return  diff;
}
于 2017-01-09T05:19:45.120 回答