在Java中有没有一种简单的方法可以做到这一点？java.util.Date类似乎相当有限。

下面是我的快速而肮脏的Java解决方案：

import java.util.Date;
import javax.management.timer.Timer;

String getRelativeDate(Date date) {     
  long delta = new Date().getTime() - date.getTime();
  if (delta < 1L * Timer.ONE_MINUTE) {
    return toSeconds(delta) == 1 ? "one second ago" : toSeconds(delta) + " seconds ago";
  }
  if (delta < 2L * Timer.ONE_MINUTE) {
    return "a minute ago";
  }
  if (delta < 45L * Timer.ONE_MINUTE) {
    return toMinutes(delta) + " minutes ago";
  }
  if (delta < 90L * Timer.ONE_MINUTE) {
    return "an hour ago";
  }
  if (delta < 24L * Timer.ONE_HOUR) {
    return toHours(delta) + " hours ago";
  }
  if (delta < 48L * Timer.ONE_HOUR) {
    return "yesterday";
  }
  if (delta < 30L * Timer.ONE_DAY) {
    return toDays(delta) + " days ago";
  }
  if (delta < 12L * 4L * Timer.ONE_WEEK) { // a month
    long months = toMonths(delta); 
    return months <= 1 ? "one month ago" : months + " months ago";
  }
  else {
    long years = toYears(delta);
    return years <= 1 ? "one year ago" : years + " years ago";
  }
}

private long toSeconds(long date) {
  return date / 1000L;
}

private long toMinutes(long date) {
  return toSeconds(date) / 60L;
}

private long toHours(long date) {
  return toMinutes(date) / 60L;
}

private long toDays(long date) {
  return toHours(date) / 24L;
}

private long toMonths(long date) {
  return toDays(date) / 30L;
}

private long toYears(long date) {
  return toMonths(date) / 365L;
}

使用Fluent DateTime

var dateTime1 = 2.Hours().Ago();
var dateTime2 = 3.Days().Ago();
var dateTime3 = 1.Months().Ago();
var dateTime4 = 5.Hours().FromNow();
var dateTime5 = 2.Weeks().FromNow();
var dateTime6 = 40.Seconds().FromNow();

在Java中有没有一种简单的方法可以做到这一点？java.util.Date类似乎相当有限。

下面是我的快速而肮脏的Java解决方案：

import java.util.Date;
import javax.management.timer.Timer;

String getRelativeDate(Date date) {     
  long delta = new Date().getTime() - date.getTime();
  if (delta < 1L * Timer.ONE_MINUTE) {
    return toSeconds(delta) == 1 ? "one second ago" : toSeconds(delta) + " seconds ago";
  }
  if (delta < 2L * Timer.ONE_MINUTE) {
    return "a minute ago";
  }
  if (delta < 45L * Timer.ONE_MINUTE) {
    return toMinutes(delta) + " minutes ago";
  }
  if (delta < 90L * Timer.ONE_MINUTE) {
    return "an hour ago";
  }
  if (delta < 24L * Timer.ONE_HOUR) {
    return toHours(delta) + " hours ago";
  }
  if (delta < 48L * Timer.ONE_HOUR) {
    return "yesterday";
  }
  if (delta < 30L * Timer.ONE_DAY) {
    return toDays(delta) + " days ago";
  }
  if (delta < 12L * 4L * Timer.ONE_WEEK) { // a month
    long months = toMonths(delta); 
    return months <= 1 ? "one month ago" : months + " months ago";
  }
  else {
    long years = toYears(delta);
    return years <= 1 ? "one year ago" : years + " years ago";
  }
}

private long toSeconds(long date) {
  return date / 1000L;
}

private long toMinutes(long date) {
  return toSeconds(date) / 60L;
}

private long toHours(long date) {
  return toMinutes(date) / 60L;
}

private long toDays(long date) {
  return toHours(date) / 24L;
}

private long toMonths(long date) {
  return toDays(date) / 30L;
}

private long toYears(long date) {
  return toMonths(date) / 365L;
}

这是stackoverflow使用的算法，但使用了错误修复（没有“一小时前”）的perlish伪代码进行了更简洁的重写。该函数在秒前取一个（正数），并返回一个人类友好的字符串，如“3小时前”或“昨天”。

agoify($delta)
  local($y, $mo, $d, $h, $m, $s);
  $s = floor($delta);
  if($s<=1)            return "a second ago";
  if($s<60)            return "$s seconds ago";
  $m = floor($s/60);
  if($m==1)            return "a minute ago";
  if($m<45)            return "$m minutes ago";
  $h = floor($m/60);
  if($h==1)            return "an hour ago";
  if($h<24)            return "$h hours ago";
  $d = floor($h/24);
  if($d<2)             return "yesterday";
  if($d<30)            return "$d days ago";
  $mo = floor($d/30);
  if($mo<=1)           return "a month ago";
  $y = floor($mo/12);
  if($y<1)             return "$mo months ago";
  if($y==1)            return "a year ago";
  return "$y years ago";

@杰夫

var ts=新时间跨度（DateTime.UtcNow.Ticks-dt.Ticks）；

对DateTime执行减法仍会返回TimeSpan。

所以你可以这样做

(DateTime.UtcNow - dt).TotalSeconds

我也很惊讶地看到常数用手相乘，然后注释加上乘法。这是错误的优化吗？

// Calculate total days in current year
int daysInYear;

for (var i = 1; i <= 12; i++)
    daysInYear += DateTime.DaysInMonth(DateTime.Now.Year, i);

// Past date
DateTime dateToCompare = DateTime.Now.Subtract(TimeSpan.FromMinutes(582));

// Calculate difference between current date and past date
double diff = (DateTime.Now - dateToCompare).TotalMilliseconds;

TimeSpan ts = TimeSpan.FromMilliseconds(diff);

var years = ts.TotalDays / daysInYear; // Years
var months = ts.TotalDays / (daysInYear / (double)12); // Months
var weeks = ts.TotalDays / 7; // Weeks
var days = ts.TotalDays; // Days
var hours = ts.TotalHours; // Hours
var minutes = ts.TotalMinutes; // Minutes
var seconds = ts.TotalSeconds; // Seconds

if (years >= 1)
    Console.WriteLine(Math.Round(years, 0) + " year(s) ago");
else if (months >= 1)
    Console.WriteLine(Math.Round(months, 0) + " month(s) ago");
else if (weeks >= 1)
    Console.WriteLine(Math.Round(weeks, 0) + " week(s) ago");
else if (days >= 1)
    Console.WriteLine(Math.Round(days, 0) + " days(s) ago");
else if (hours >= 1)
    Console.WriteLine(Math.Round(hours, 0) + " hour(s) ago");
else if (minutes >= 1)
    Console.WriteLine(Math.Round(minutes, 0) + " minute(s) ago");
else if (seconds >= 1)
    Console.WriteLine(Math.Round(seconds, 0) + " second(s) ago");

Console.ReadLine();