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
聚会晚了几年,但我有一个要求,无论是过去还是将来的约会,都要这样做,所以我把杰夫和文森特的约会结合在一起。这是一场盛大的盛会!:)
public static class DateTimeHelper
{
private const int SECOND = 1;
private const int MINUTE = 60 * SECOND;
private const int HOUR = 60 * MINUTE;
private const int DAY = 24 * HOUR;
private const int MONTH = 30 * DAY;
/// <summary>
/// Returns a friendly version of the provided DateTime, relative to now. E.g.: "2 days ago", or "in 6 months".
/// </summary>
/// <param name="dateTime">The DateTime to compare to Now</param>
/// <returns>A friendly string</returns>
public static string GetFriendlyRelativeTime(DateTime dateTime)
{
if (DateTime.UtcNow.Ticks == dateTime.Ticks)
{
return "Right now!";
}
bool isFuture = (DateTime.UtcNow.Ticks < dateTime.Ticks);
var ts = DateTime.UtcNow.Ticks < dateTime.Ticks ? new TimeSpan(dateTime.Ticks - DateTime.UtcNow.Ticks) : new TimeSpan(DateTime.UtcNow.Ticks - dateTime.Ticks);
double delta = ts.TotalSeconds;
if (delta < 1 * MINUTE)
{
return isFuture ? "in " + (ts.Seconds == 1 ? "one second" : ts.Seconds + " seconds") : ts.Seconds == 1 ? "one second ago" : ts.Seconds + " seconds ago";
}
if (delta < 2 * MINUTE)
{
return isFuture ? "in a minute" : "a minute ago";
}
if (delta < 45 * MINUTE)
{
return isFuture ? "in " + ts.Minutes + " minutes" : ts.Minutes + " minutes ago";
}
if (delta < 90 * MINUTE)
{
return isFuture ? "in an hour" : "an hour ago";
}
if (delta < 24 * HOUR)
{
return isFuture ? "in " + ts.Hours + " hours" : ts.Hours + " hours ago";
}
if (delta < 48 * HOUR)
{
return isFuture ? "tomorrow" : "yesterday";
}
if (delta < 30 * DAY)
{
return isFuture ? "in " + ts.Days + " days" : ts.Days + " days ago";
}
if (delta < 12 * MONTH)
{
int months = Convert.ToInt32(Math.Floor((double)ts.Days / 30));
return isFuture ? "in " + (months <= 1 ? "one month" : months + " months") : months <= 1 ? "one month ago" : months + " months ago";
}
else
{
int years = Convert.ToInt32(Math.Floor((double)ts.Days / 365));
return isFuture ? "in " + (years <= 1 ? "one year" : years + " years") : years <= 1 ? "one year ago" : years + " years ago";
}
}
}
其他回答
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。
using System;
using System.Collections.Generic;
using System.Linq;
public static class RelativeDateHelper
{
private static Dictionary<double, Func<double, string>> sm_Dict = null;
private static Dictionary<double, Func<double, string>> DictionarySetup()
{
var dict = new Dictionary<double, Func<double, string>>();
dict.Add(0.75, (mins) => "less than a minute");
dict.Add(1.5, (mins) => "about a minute");
dict.Add(45, (mins) => string.Format("{0} minutes", Math.Round(mins)));
dict.Add(90, (mins) => "about an hour");
dict.Add(1440, (mins) => string.Format("about {0} hours", Math.Round(Math.Abs(mins / 60)))); // 60 * 24
dict.Add(2880, (mins) => "a day"); // 60 * 48
dict.Add(43200, (mins) => string.Format("{0} days", Math.Floor(Math.Abs(mins / 1440)))); // 60 * 24 * 30
dict.Add(86400, (mins) => "about a month"); // 60 * 24 * 60
dict.Add(525600, (mins) => string.Format("{0} months", Math.Floor(Math.Abs(mins / 43200)))); // 60 * 24 * 365
dict.Add(1051200, (mins) => "about a year"); // 60 * 24 * 365 * 2
dict.Add(double.MaxValue, (mins) => string.Format("{0} years", Math.Floor(Math.Abs(mins / 525600))));
return dict;
}
public static string ToRelativeDate(this DateTime input)
{
TimeSpan oSpan = DateTime.Now.Subtract(input);
double TotalMinutes = oSpan.TotalMinutes;
string Suffix = " ago";
if (TotalMinutes < 0.0)
{
TotalMinutes = Math.Abs(TotalMinutes);
Suffix = " from now";
}
if (null == sm_Dict)
sm_Dict = DictionarySetup();
return sm_Dict.First(n => TotalMinutes < n.Key).Value.Invoke(TotalMinutes) + Suffix;
}
}
与此问题的另一个答案相同,但作为静态字典的扩展方法。
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;
}
iPhone Objective-C版本
+ (NSString *)timeAgoString:(NSDate *)date {
int delta = -(int)[date timeIntervalSinceNow];
if (delta < 60)
{
return delta == 1 ? @"one second ago" : [NSString stringWithFormat:@"%i seconds ago", delta];
}
if (delta < 120)
{
return @"a minute ago";
}
if (delta < 2700)
{
return [NSString stringWithFormat:@"%i minutes ago", delta/60];
}
if (delta < 5400)
{
return @"an hour ago";
}
if (delta < 24 * 3600)
{
return [NSString stringWithFormat:@"%i hours ago", delta/3600];
}
if (delta < 48 * 3600)
{
return @"yesterday";
}
if (delta < 30 * 24 * 3600)
{
return [NSString stringWithFormat:@"%i days ago", delta/(24*3600)];
}
if (delta < 12 * 30 * 24 * 3600)
{
int months = delta/(30*24*3600);
return months <= 1 ? @"one month ago" : [NSString stringWithFormat:@"%i months ago", months];
}
else
{
int years = delta/(12*30*24*3600);
return years <= 1 ? @"one year ago" : [NSString stringWithFormat:@"%i years ago", years];
}
}
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";