给定特定的DateTime值,如何显示相对时间,例如:

2小时前3天前一个月前


当前回答

以某种方式,您可以使用DateTime函数以秒到年计算相对时间,请尝试以下操作:

using System;

public class Program {
    public static string getRelativeTime(DateTime past) {
        DateTime now = DateTime.Today;
        string rt = "";
        int time;
        string statement = "";
        if (past.Second >= now.Second) {
            if (past.Second - now.Second == 1) {
                rt = "second ago";
            }
            rt = "seconds ago";
            time = past.Second - now.Second;
            statement = "" + time;
            return (statement + rt);
        }
        if (past.Minute >= now.Minute) {
            if (past.Second - now.Second == 1) {
                rt = "second ago";
            } else {
                rt = "minutes ago";
            }
            time = past.Minute - now.Minute;
            statement = "" + time;
            return (statement + rt);
        }
        // This process will go on until years
    }
    public static void Main() {
        DateTime before = new DateTime(1995, 8, 24);
        string date = getRelativeTime(before);
        Console.WriteLine("Windows 95 was {0}.", date);
    }
}

不完全有效,但如果您对其进行一点修改和调试,它很可能会完成任务。

其他回答

文斯回答的土耳其语本地化版本。

    const int SECOND = 1;
    const int MINUTE = 60 * SECOND;
    const int HOUR = 60 * MINUTE;
    const int DAY = 24 * HOUR;
    const int MONTH = 30 * DAY;

    var ts = new TimeSpan(DateTime.UtcNow.Ticks - yourDate.Ticks);
    double delta = Math.Abs(ts.TotalSeconds);

    if (delta < 1 * MINUTE)
        return ts.Seconds + " saniye önce";

    if (delta < 45 * MINUTE)
        return ts.Minutes + " dakika önce";

    if (delta < 24 * HOUR)
        return ts.Hours + " saat önce";

    if (delta < 48 * HOUR)
        return "dün";

    if (delta < 30 * DAY)
        return ts.Days + " gün önce";

    if (delta < 12 * MONTH)
    {
        int months = Convert.ToInt32(Math.Floor((double)ts.Days / 30));
        return months + " ay önce";
    }
    else
    {
        int years = Convert.ToInt32(Math.Floor((double)ts.Days / 365));
        return years + " yıl önce";
    }

这里是Jeffs Script for PHP的重写:

define("SECOND", 1);
define("MINUTE", 60 * SECOND);
define("HOUR", 60 * MINUTE);
define("DAY", 24 * HOUR);
define("MONTH", 30 * DAY);
function relativeTime($time)
{   
    $delta = time() - $time;

    if ($delta < 1 * MINUTE)
    {
        return $delta == 1 ? "one second ago" : $delta . " seconds ago";
    }
    if ($delta < 2 * MINUTE)
    {
      return "a minute ago";
    }
    if ($delta < 45 * MINUTE)
    {
        return floor($delta / MINUTE) . " minutes ago";
    }
    if ($delta < 90 * MINUTE)
    {
      return "an hour ago";
    }
    if ($delta < 24 * HOUR)
    {
      return floor($delta / HOUR) . " hours ago";
    }
    if ($delta < 48 * HOUR)
    {
      return "yesterday";
    }
    if ($delta < 30 * DAY)
    {
        return floor($delta / DAY) . " days ago";
    }
    if ($delta < 12 * MONTH)
    {
      $months = floor($delta / DAY / 30);
      return $months <= 1 ? "one month ago" : $months . " months ago";
    }
    else
    {
        $years = floor($delta / DAY / 365);
        return $years <= 1 ? "one year ago" : $years . " years ago";
    }
}    

当然,解决“1小时前”问题的一个简单方法是增加“一小时前”有效的窗口。改变

if (delta < 5400) // 90 * 60
{
    return "an hour ago";
}

into

if (delta < 7200) // 120 * 60
{
    return "an hour ago";
}

这意味着110分钟前发生的事情将被解读为“一小时前”——这可能并不完美,但我认为这比“1小时前”的现状要好。

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";

如果您想获得类似“2天4小时12分钟前”的输出,则需要一个时间跨度:

TimeSpan timeDiff = DateTime.Now-CreatedDate;

然后您可以访问您喜欢的值:

timeDiff.Days
timeDiff.Hours