我认为已经有很多关于这篇文章的答案了，但你可以使用它，它就像插件一样容易使用，程序员也很容易阅读。发送您的特定日期，并以字符串形式获取其值：

public string RelativeDateTimeCount(DateTime inputDateTime)
{
    string outputDateTime = string.Empty;
    TimeSpan ts = DateTime.Now - inputDateTime;

    if (ts.Days > 7)
    { outputDateTime = inputDateTime.ToString("MMMM d, yyyy"); }

    else if (ts.Days > 0)
    {
        outputDateTime = ts.Days == 1 ? ("about 1 Day ago") : ("about " + ts.Days.ToString() + " Days ago");
    }
    else if (ts.Hours > 0)
    {
        outputDateTime = ts.Hours == 1 ? ("an hour ago") : (ts.Hours.ToString() + " hours ago");
    }
    else if (ts.Minutes > 0)
    {
        outputDateTime = ts.Minutes == 1 ? ("1 minute ago") : (ts.Minutes.ToString() + " minutes ago");
    }
    else outputDateTime = "few seconds ago";

    return outputDateTime;
}

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;
}

简单且100%的工作解决方案。

处理过去和将来的时间。。以防万一

        public string GetTimeSince(DateTime postDate)
    {
        string message = "";
        DateTime currentDate = DateTime.Now;
        TimeSpan timegap = currentDate - postDate;

     
        if (timegap.Days > 365)
        {
            message = string.Format(L("Ago") + " {0} " + L("Years"), (((timegap.Days) / 30) / 12));                
        }
        else if (timegap.Days > 30)
        {
            message = string.Format(L("Ago") + " {0} " + L("Months"), timegap.Days/30);                
        }
        else if (timegap.Days > 0)
        {
            message = string.Format(L("Ago") + " {0} " + L("Days"), timegap.Days);
        }           
        else if (timegap.Hours > 0)
        {
            message = string.Format(L("Ago") + " {0} " + L("Hours"), timegap.Hours);
        }           
        else if (timegap.Minutes > 0)
        {
            message = string.Format(L("Ago") + " {0} " + L("Minutes"), timegap.Minutes);
        }
        else if (timegap.Seconds > 0)
        {
            message = string.Format(L("Ago") + " {0} " + L("Seconds"), timegap.Seconds);
        }

        // let's handle future times..just in case       
        else if (timegap.Days < -365)
        {
            message = string.Format(L("In") + " {0} " + L("Years"), (((Math.Abs(timegap.Days)) / 30) / 12));                
        }
        else if (timegap.Days < -30)
        {
            message = string.Format(L("In") + " {0} " + L("Months"), ((Math.Abs(timegap.Days)) / 30));                
        }
        else if (timegap.Days < 0)
        {
            message = string.Format(L("In") + " {0} " + L("Days"), Math.Abs(timegap.Days));                
        }           
      
        else if (timegap.Hours < 0)
        {
            message = string.Format(L("In") + " {0} " + L("Hours"), Math.Abs(timegap.Hours));                
        }
        else if (timegap.Minutes < 0)
        {
            message = string.Format(L("In") + " {0} " + L("Minutes"), Math.Abs(timegap.Minutes));                
        }
        else if (timegap.Seconds < 0)
        {
            message = string.Format(L("In") + " {0} " + L("Seconds"), Math.Abs(timegap.Seconds));                
        }


        else
        {
            message = "a bit";
        }

        return message;
    }

Nuget上还有一个名为Humanizr的软件包，它实际上运行得很好，并且在.NET Foundation中。

DateTime.UtcNow.AddHours(-30).Humanize() => "yesterday"
DateTime.UtcNow.AddHours(-2).Humanize() => "2 hours ago"

DateTime.UtcNow.AddHours(30).Humanize() => "tomorrow"
DateTime.UtcNow.AddHours(2).Humanize() => "2 hours from now"

TimeSpan.FromMilliseconds(1299630020).Humanize() => "2 weeks"
TimeSpan.FromMilliseconds(1299630020).Humanize(3) => "2 weeks, 1 day, 1 hour"

Scott Hanselman在他的博客上写了一篇文章

@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？