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
在PHP中,我是这样做的:
<?php
function timesince($original) {
// array of time period chunks
$chunks = array(
array(60 * 60 * 24 * 365 , 'year'),
array(60 * 60 * 24 * 30 , 'month'),
array(60 * 60 * 24 * 7, 'week'),
array(60 * 60 * 24 , 'day'),
array(60 * 60 , 'hour'),
array(60 , 'minute'),
);
$today = time(); /* Current unix time */
$since = $today - $original;
if($since > 604800) {
$print = date("M jS", $original);
if($since > 31536000) {
$print .= ", " . date("Y", $original);
}
return $print;
}
// $j saves performing the count function each time around the loop
for ($i = 0, $j = count($chunks); $i < $j; $i++) {
$seconds = $chunks[$i][0];
$name = $chunks[$i][1];
// finding the biggest chunk (if the chunk fits, break)
if (($count = floor($since / $seconds)) != 0) {
break;
}
}
$print = ($count == 1) ? '1 '.$name : "$count {$name}s";
return $print . " ago";
} ?>
其他回答
使用解构主义和Linq得到“n(最大时间单位)前”的“一行”:
TimeSpan timeSpan = DateTime.Now - new DateTime(1234, 5, 6, 7, 8, 9);
(string unit, int value) = new Dictionary<string, int>
{
{"year(s)", (int)(timeSpan.TotalDays / 365.25)}, //https://en.wikipedia.org/wiki/Year#Intercalation
{"month(s)", (int)(timeSpan.TotalDays / 29.53)}, //https://en.wikipedia.org/wiki/Month
{"day(s)", (int)timeSpan.TotalDays},
{"hour(s)", (int)timeSpan.TotalHours},
{"minute(s)", (int)timeSpan.TotalMinutes},
{"second(s)", (int)timeSpan.TotalSeconds},
{"millisecond(s)", (int)timeSpan.TotalMilliseconds}
}.First(kvp => kvp.Value > 0);
Console.WriteLine($"{value} {unit} ago");
你在786年前
当前年份和月份,如
TimeSpan timeSpan = DateTime.Now - new DateTime(2020, 12, 6, 7, 8, 9);
您4天前收到
实际日期,比如
TimeSpan timeSpan = DateTime.Now - DateTime.Now.Date;
9小时前到达
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。
public static string ToRelativeDate(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";
}
var aValue = new SortedList<double, Func<string>>();
aValue.Add(0.75, () => "less than a minute");
aValue.Add(1.5, () => "about a minute");
aValue.Add(45, () => string.Format("{0} minutes", Math.Round(TotalMinutes)));
aValue.Add(90, () => "about an hour");
aValue.Add(1440, () => string.Format("about {0} hours", Math.Round(Math.Abs(oSpan.TotalHours)))); // 60 * 24
aValue.Add(2880, () => "a day"); // 60 * 48
aValue.Add(43200, () => string.Format("{0} days", Math.Floor(Math.Abs(oSpan.TotalDays)))); // 60 * 24 * 30
aValue.Add(86400, () => "about a month"); // 60 * 24 * 60
aValue.Add(525600, () => string.Format("{0} months", Math.Floor(Math.Abs(oSpan.TotalDays / 30)))); // 60 * 24 * 365
aValue.Add(1051200, () => "about a year"); // 60 * 24 * 365 * 2
aValue.Add(double.MaxValue, () => string.Format("{0} years", Math.Floor(Math.Abs(oSpan.TotalDays / 365))));
return aValue.First(n => TotalMinutes < n.Key).Value.Invoke() + Suffix;
}
http://refactormycode.com/codes/493-twitter-esque-relative-dates
C#6版本:
static readonly SortedList<double, Func<TimeSpan, string>> offsets =
new SortedList<double, Func<TimeSpan, string>>
{
{ 0.75, _ => "less than a minute"},
{ 1.5, _ => "about a minute"},
{ 45, x => $"{x.TotalMinutes:F0} minutes"},
{ 90, x => "about an hour"},
{ 1440, x => $"about {x.TotalHours:F0} hours"},
{ 2880, x => "a day"},
{ 43200, x => $"{x.TotalDays:F0} days"},
{ 86400, x => "about a month"},
{ 525600, x => $"{x.TotalDays / 30:F0} months"},
{ 1051200, x => "about a year"},
{ double.MaxValue, x => $"{x.TotalDays / 365:F0} years"}
};
public static string ToRelativeDate(this DateTime input)
{
TimeSpan x = DateTime.Now - input;
string Suffix = x.TotalMinutes > 0 ? " ago" : " from now";
x = new TimeSpan(Math.Abs(x.Ticks));
return offsets.First(n => x.TotalMinutes < n.Key).Value(x) + Suffix;
}
我是这样做的
var ts = new TimeSpan(DateTime.UtcNow.Ticks - dt.Ticks);
double delta = Math.Abs(ts.TotalSeconds);
if (delta < 60)
{
return ts.Seconds == 1 ? "one second ago" : ts.Seconds + " seconds ago";
}
if (delta < 60 * 2)
{
return "a minute ago";
}
if (delta < 45 * 60)
{
return ts.Minutes + " minutes ago";
}
if (delta < 90 * 60)
{
return "an hour ago";
}
if (delta < 24 * 60 * 60)
{
return ts.Hours + " hours ago";
}
if (delta < 48 * 60 * 60)
{
return "yesterday";
}
if (delta < 30 * 24 * 60 * 60)
{
return ts.Days + " days ago";
}
if (delta < 12 * 30 * 24 * 60 * 60)
{
int months = Convert.ToInt32(Math.Floor((double)ts.Days / 30));
return months <= 1 ? "one month ago" : months + " months ago";
}
int years = Convert.ToInt32(Math.Floor((double)ts.Days / 365));
return years <= 1 ? "one year ago" : years + " years ago";
建议?评论?如何改进此算法?
这是stackoverflow使用的算法,但使用了错误修复(没有“一小时前”)的perlish伪代码进行了更简洁的重写。该函数在秒前取一个(正数),并返回一个人类友好的字符串,如“3小时前”或“昨天”。
agoify($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";