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
下面是我作为DateTime类的扩展方法添加的一个实现,它处理未来和过去的日期,并提供了一个近似选项,允许您指定要查找的详细程度(“3小时前”与“3小时、23分钟、12秒前”):
using System.Text;
/// <summary>
/// Compares a supplied date to the current date and generates a friendly English
/// comparison ("5 days ago", "5 days from now")
/// </summary>
/// <param name="date">The date to convert</param>
/// <param name="approximate">When off, calculate timespan down to the second.
/// When on, approximate to the largest round unit of time.</param>
/// <returns></returns>
public static string ToRelativeDateString(this DateTime value, bool approximate)
{
StringBuilder sb = new StringBuilder();
string suffix = (value > DateTime.Now) ? " from now" : " ago";
TimeSpan timeSpan = new TimeSpan(Math.Abs(DateTime.Now.Subtract(value).Ticks));
if (timeSpan.Days > 0)
{
sb.AppendFormat("{0} {1}", timeSpan.Days,
(timeSpan.Days > 1) ? "days" : "day");
if (approximate) return sb.ToString() + suffix;
}
if (timeSpan.Hours > 0)
{
sb.AppendFormat("{0}{1} {2}", (sb.Length > 0) ? ", " : string.Empty,
timeSpan.Hours, (timeSpan.Hours > 1) ? "hours" : "hour");
if (approximate) return sb.ToString() + suffix;
}
if (timeSpan.Minutes > 0)
{
sb.AppendFormat("{0}{1} {2}", (sb.Length > 0) ? ", " : string.Empty,
timeSpan.Minutes, (timeSpan.Minutes > 1) ? "minutes" : "minute");
if (approximate) return sb.ToString() + suffix;
}
if (timeSpan.Seconds > 0)
{
sb.AppendFormat("{0}{1} {2}", (sb.Length > 0) ? ", " : string.Empty,
timeSpan.Seconds, (timeSpan.Seconds > 1) ? "seconds" : "second");
if (approximate) return sb.ToString() + suffix;
}
if (sb.Length == 0) return "right now";
sb.Append(suffix);
return sb.ToString();
}
其他回答
下面是我作为DateTime类的扩展方法添加的一个实现,它处理未来和过去的日期,并提供了一个近似选项,允许您指定要查找的详细程度(“3小时前”与“3小时、23分钟、12秒前”):
using System.Text;
/// <summary>
/// Compares a supplied date to the current date and generates a friendly English
/// comparison ("5 days ago", "5 days from now")
/// </summary>
/// <param name="date">The date to convert</param>
/// <param name="approximate">When off, calculate timespan down to the second.
/// When on, approximate to the largest round unit of time.</param>
/// <returns></returns>
public static string ToRelativeDateString(this DateTime value, bool approximate)
{
StringBuilder sb = new StringBuilder();
string suffix = (value > DateTime.Now) ? " from now" : " ago";
TimeSpan timeSpan = new TimeSpan(Math.Abs(DateTime.Now.Subtract(value).Ticks));
if (timeSpan.Days > 0)
{
sb.AppendFormat("{0} {1}", timeSpan.Days,
(timeSpan.Days > 1) ? "days" : "day");
if (approximate) return sb.ToString() + suffix;
}
if (timeSpan.Hours > 0)
{
sb.AppendFormat("{0}{1} {2}", (sb.Length > 0) ? ", " : string.Empty,
timeSpan.Hours, (timeSpan.Hours > 1) ? "hours" : "hour");
if (approximate) return sb.ToString() + suffix;
}
if (timeSpan.Minutes > 0)
{
sb.AppendFormat("{0}{1} {2}", (sb.Length > 0) ? ", " : string.Empty,
timeSpan.Minutes, (timeSpan.Minutes > 1) ? "minutes" : "minute");
if (approximate) return sb.ToString() + suffix;
}
if (timeSpan.Seconds > 0)
{
sb.AppendFormat("{0}{1} {2}", (sb.Length > 0) ? ", " : string.Empty,
timeSpan.Seconds, (timeSpan.Seconds > 1) ? "seconds" : "second");
if (approximate) return sb.ToString() + suffix;
}
if (sb.Length == 0) return "right now";
sb.Append(suffix);
return sb.ToString();
}
Jeff,您的代码很好,但使用常量可以更清晰(如代码完成中所建议的)。
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 == 1 ? "one second ago" : ts.Seconds + " seconds ago";
if (delta < 2 * MINUTE)
return "a minute ago";
if (delta < 45 * MINUTE)
return ts.Minutes + " minutes ago";
if (delta < 90 * MINUTE)
return "an hour ago";
if (delta < 24 * HOUR)
return ts.Hours + " hours ago";
if (delta < 48 * HOUR)
return "yesterday";
if (delta < 30 * DAY)
return ts.Days + " days ago";
if (delta < 12 * MONTH)
{
int months = Convert.ToInt32(Math.Floor((double)ts.Days / 30));
return months <= 1 ? "one month ago" : months + " months ago";
}
else
{
int years = Convert.ToInt32(Math.Floor((double)ts.Days / 365));
return years <= 1 ? "one year ago" : years + " years ago";
}
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。
当您知道查看者的时区时,以日为单位使用日历日可能会更清晰。我不熟悉.NET库,所以我不知道如何在C#中实现这一点。
在消费者网站上,你也可以在一分钟内用手洗。“不到一分钟前”或“刚刚”就足够了。
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";
}
}