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
public static string RelativeDate(DateTime theDate)
{
Dictionary<long, string> thresholds = new Dictionary<long, string>();
int minute = 60;
int hour = 60 * minute;
int day = 24 * hour;
thresholds.Add(60, "{0} seconds ago");
thresholds.Add(minute * 2, "a minute ago");
thresholds.Add(45 * minute, "{0} minutes ago");
thresholds.Add(120 * minute, "an hour ago");
thresholds.Add(day, "{0} hours ago");
thresholds.Add(day * 2, "yesterday");
thresholds.Add(day * 30, "{0} days ago");
thresholds.Add(day * 365, "{0} months ago");
thresholds.Add(long.MaxValue, "{0} years ago");
long since = (DateTime.Now.Ticks - theDate.Ticks) / 10000000;
foreach (long threshold in thresholds.Keys)
{
if (since < threshold)
{
TimeSpan t = new TimeSpan((DateTime.Now.Ticks - theDate.Ticks));
return string.Format(thresholds[threshold], (t.Days > 365 ? t.Days / 365 : (t.Days > 0 ? t.Days : (t.Hours > 0 ? t.Hours : (t.Minutes > 0 ? t.Minutes : (t.Seconds > 0 ? t.Seconds : 0))))).ToString());
}
}
return "";
}
我更喜欢这个版本,因为它简洁,并且能够添加新的刻度点。这可以用Timespan的Latest()扩展来封装,而不是长的1行,但为了发布的简洁,这可以。这修复了一小时前、一小时前的问题,提供了一个小时直到两小时过去
其他回答
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";
}
}
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();
}
我是这样做的
var ts = new TimeSpan(DateTime.UtcNow.Ticks - dt.Ticks);
double delta = Math.Abs(ts.TotalSeconds);
if (delta < 60)
{
return ts.Seconds == 1 ? "one second ago" : ts.Seconds + " seconds ago";
}
if (delta < 60 * 2)
{
return "a minute ago";
}
if (delta < 45 * 60)
{
return ts.Minutes + " minutes ago";
}
if (delta < 90 * 60)
{
return "an hour ago";
}
if (delta < 24 * 60 * 60)
{
return ts.Hours + " hours ago";
}
if (delta < 48 * 60 * 60)
{
return "yesterday";
}
if (delta < 30 * 24 * 60 * 60)
{
return ts.Days + " days ago";
}
if (delta < 12 * 30 * 24 * 60 * 60)
{
int months = Convert.ToInt32(Math.Floor((double)ts.Days / 30));
return months <= 1 ? "one month ago" : months + " months ago";
}
int years = Convert.ToInt32(Math.Floor((double)ts.Days / 365));
return years <= 1 ? "one year ago" : years + " years ago";
建议?评论?如何改进此算法?
我的方法要简单得多。您可以根据需要调整返回字符串
public static string TimeLeft(DateTime utcDate)
{
TimeSpan timeLeft = DateTime.UtcNow - utcDate;
string timeLeftString = "";
if (timeLeft.Days > 0)
{
timeLeftString += timeLeft.Days == 1 ? timeLeft.Days + " day" : timeLeft.Days + " days";
}
else if (timeLeft.Hours > 0)
{
timeLeftString += timeLeft.Hours == 1 ? timeLeft.Hours + " hour" : timeLeft.Hours + " hours";
}
else
{
timeLeftString += timeLeft.Minutes == 1 ? timeLeft.Minutes+" minute" : timeLeft.Minutes + " minutes";
}
return timeLeftString;
}
@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?