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
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
等