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
我也建议在客户端进行计算。服务器工作更少。
以下是我使用的版本(来自Zach Leatherman)
/*
* Javascript Humane Dates
* Copyright (c) 2008 Dean Landolt (deanlandolt.com)
* Re-write by Zach Leatherman (zachleat.com)
*
* Adopted from the John Resig's pretty.js
* at http://ejohn.org/blog/javascript-pretty-date
* and henrah's proposed modification
* at http://ejohn.org/blog/javascript-pretty-date/#comment-297458
*
* Licensed under the MIT license.
*/
function humane_date(date_str){
var time_formats = [
[60, 'just now'],
[90, '1 minute'], // 60*1.5
[3600, 'minutes', 60], // 60*60, 60
[5400, '1 hour'], // 60*60*1.5
[86400, 'hours', 3600], // 60*60*24, 60*60
[129600, '1 day'], // 60*60*24*1.5
[604800, 'days', 86400], // 60*60*24*7, 60*60*24
[907200, '1 week'], // 60*60*24*7*1.5
[2628000, 'weeks', 604800], // 60*60*24*(365/12), 60*60*24*7
[3942000, '1 month'], // 60*60*24*(365/12)*1.5
[31536000, 'months', 2628000], // 60*60*24*365, 60*60*24*(365/12)
[47304000, '1 year'], // 60*60*24*365*1.5
[3153600000, 'years', 31536000], // 60*60*24*365*100, 60*60*24*365
[4730400000, '1 century'] // 60*60*24*365*100*1.5
];
var time = ('' + date_str).replace(/-/g,"/").replace(/[TZ]/g," "),
dt = new Date,
seconds = ((dt - new Date(time) + (dt.getTimezoneOffset() * 60000)) / 1000),
token = ' ago',
i = 0,
format;
if (seconds < 0) {
seconds = Math.abs(seconds);
token = '';
}
while (format = time_formats[i++]) {
if (seconds < format[0]) {
if (format.length == 2) {
return format[1] + (i > 1 ? token : ''); // Conditional so we don't return Just Now Ago
} else {
return Math.round(seconds / format[2]) + ' ' + format[1] + (i > 1 ? token : '');
}
}
}
// overflow for centuries
if(seconds > 4730400000)
return Math.round(seconds / 4730400000) + ' centuries' + token;
return date_str;
};
if(typeof jQuery != 'undefined') {
jQuery.fn.humane_dates = function(){
return this.each(function(){
var date = humane_date(this.title);
if(date && jQuery(this).text() != date) // don't modify the dom if we don't have to
jQuery(this).text(date);
});
};
}
@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?
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?
/**
* {@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";
} ?>