下面的代码可以给你想要的输出:

String startDate = "Jan 01 2015";
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("MMM dd yyyy");
LocalDate date = LocalDate.parse(startDate, formatter);

String currentDate = "Feb 11 2015";
LocalDate date1 = LocalDate.parse(currentDate, formatter);

System.out.println(date1.toEpochDay() - date.toEpochDay());

int daysDiff = (date1.getTime() - date2.getTime()) / MILLIS_PER_DAY;

因为这个问题用Scala做了标记，

import scala.concurrent.duration._
val diff = (System.currentTimeMillis() - oldDate.getTime).milliseconds
val diffSeconds = diff.toSeconds
val diffMinutes = diff.toMinutes
val diffHours = diff.toHours
val diffDays = diff.toDays

试试这个:

int epoch = (int) (new java.text.SimpleDateFormat("MM/dd/yyyy HH:mm:ss").parse("01/01/1970  00:00:00").getTime() / 1000);

你可以在parse()方法参数中编辑字符串。

不使用标准API，不行。你可以这样做:

class Duration {
    private final TimeUnit unit;
    private final long length;
    // ...
}

或者你可以使用Joda:

DateTime a = ..., b = ...;
Duration d = new Duration(a, b);

简单的diff(不含lib)

/**
 * Get a diff between two dates
 * @param date1 the oldest date
 * @param date2 the newest date
 * @param timeUnit the unit in which you want the diff
 * @return the diff value, in the provided unit
 */
public static long getDateDiff(Date date1, Date date2, TimeUnit timeUnit) {
    long diffInMillies = date2.getTime() - date1.getTime();
    return timeUnit.convert(diffInMillies,TimeUnit.MILLISECONDS);
}

然后你可以调用:

getDateDiff(date1,date2,TimeUnit.MINUTES);

以分钟为单位获取两个日期的差值。

TimeUnit是java.util.concurrent。TimeUnit，一个从纳米到天的标准Java枚举。

人类可读的差异(不含lib)

public static Map<TimeUnit,Long> computeDiff(Date date1, Date date2) {

    long diffInMillies = date2.getTime() - date1.getTime();

    //create the list
    List<TimeUnit> units = new ArrayList<TimeUnit>(EnumSet.allOf(TimeUnit.class));
    Collections.reverse(units);

    //create the result map of TimeUnit and difference
    Map<TimeUnit,Long> result = new LinkedHashMap<TimeUnit,Long>();
    long milliesRest = diffInMillies;

    for ( TimeUnit unit : units ) {
        
        //calculate difference in millisecond 
        long diff = unit.convert(milliesRest,TimeUnit.MILLISECONDS);
        long diffInMilliesForUnit = unit.toMillis(diff);
        milliesRest = milliesRest - diffInMilliesForUnit;

        //put the result in the map
        result.put(unit,diff);
    }

    return result;
}

http://ideone.com/5dXeu6

输出类似Map:{DAYS=1, HOURS=3, MINUTES=46, SECONDS=40, MILLISECONDS=0, MICROSECONDS=0, NANOSECONDS=0}，单位是有序的。

您只需将该映射转换为用户友好的字符串。

警告

上面的代码段计算两个瞬间之间的简单差。它会在夏令时切换期间导致问题，就像这篇文章中解释的那样。这意味着如果你计算没有时间的日期之间的差异，你可能会少了一天/小时。

在我看来，日期的差异是主观的，尤其是在日子上。你可以:

计算经过24小时的时间:day+1 - day = 1 day = 24h 计算经过的时间，考虑到夏令时:天+1 -天= 1 = 24小时(但使用午夜时间和夏令时，它可以是0天和23小时) 计算日开关的数量，这意味着一天+1 1pm -一天11am = 1天，即使经过的时间只有2h(如果有夏令时则为1h:p)

我的答案是有效的，如果你的日期差异的定义天匹配第一种情况

与JodaTime

如果你正在使用JodaTime，你可以得到2个瞬间的差异(millies支持ReadableInstant)日期:

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

但是你也可以得到本地日期/时间的差异:

// returns 4 because of the leap year of 366 days
new Period(LocalDate.now(), LocalDate.now().plusDays(365*5), PeriodType.years()).getYears() 

// this time it returns 5
new Period(LocalDate.now(), LocalDate.now().plusDays(365*5+1), PeriodType.years()).getYears() 

// And you can also use these static methods
Years.yearsBetween(LocalDate.now(), LocalDate.now().plusDays(365*5)).getYears()