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
我从比尔·盖茨的一个博客中得到了这个答案。我需要在我的浏览器历史记录中找到它,我会给你链接。
执行相同操作的Javascript代码(按要求):
function posted(t) {
var now = new Date();
var diff = parseInt((now.getTime() - Date.parse(t)) / 1000);
if (diff < 60) { return 'less than a minute ago'; }
else if (diff < 120) { return 'about a minute ago'; }
else if (diff < (2700)) { return (parseInt(diff / 60)).toString() + ' minutes ago'; }
else if (diff < (5400)) { return 'about an hour ago'; }
else if (diff < (86400)) { return 'about ' + (parseInt(diff / 3600)).toString() + ' hours ago'; }
else if (diff < (172800)) { return '1 day ago'; }
else {return (parseInt(diff / 86400)).toString() + ' days ago'; }
}
基本上,你是以秒为单位工作的。
其他回答
Nuget上还有一个名为Humanizr的软件包,它实际上运行得很好,并且在.NET Foundation中。
DateTime.UtcNow.AddHours(-30).Humanize() => "yesterday"
DateTime.UtcNow.AddHours(-2).Humanize() => "2 hours ago"
DateTime.UtcNow.AddHours(30).Humanize() => "tomorrow"
DateTime.UtcNow.AddHours(2).Humanize() => "2 hours from now"
TimeSpan.FromMilliseconds(1299630020).Humanize() => "2 weeks"
TimeSpan.FromMilliseconds(1299630020).Humanize(3) => "2 weeks, 1 day, 1 hour"
Scott Hanselman在他的博客上写了一篇文章
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;
}
/**
* {@code date1} has to be earlier than {@code date2}.
*/
public static String relativize(Date date1, Date date2) {
assert date2.getTime() >= date1.getTime();
long duration = date2.getTime() - date1.getTime();
long converted;
if ((converted = TimeUnit.MILLISECONDS.toDays(duration)) > 0) {
return String.format("%d %s ago", converted, converted == 1 ? "day" : "days");
} else if ((converted = TimeUnit.MILLISECONDS.toHours(duration)) > 0) {
return String.format("%d %s ago", converted, converted == 1 ? "hour" : "hours");
} else if ((converted = TimeUnit.MILLISECONDS.toMinutes(duration)) > 0) {
return String.format("%d %s ago", converted, converted == 1 ? "minute" : "minutes");
} else if ((converted = TimeUnit.MILLISECONDS.toSeconds(duration)) > 0) {
return String.format("%d %s ago", converted, converted == 1 ? "second" : "seconds");
} else {
return "just now";
}
}
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";
} ?>