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
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;
}
@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?
用于客户端gwt的Java:
import java.util.Date;
public class RelativeDateFormat {
private static final long ONE_MINUTE = 60000L;
private static final long ONE_HOUR = 3600000L;
private static final long ONE_DAY = 86400000L;
private static final long ONE_WEEK = 604800000L;
public static String format(Date date) {
long delta = new Date().getTime() - date.getTime();
if (delta < 1L * ONE_MINUTE) {
return toSeconds(delta) == 1 ? "one second ago" : toSeconds(delta)
+ " seconds ago";
}
if (delta < 2L * ONE_MINUTE) {
return "one minute ago";
}
if (delta < 45L * ONE_MINUTE) {
return toMinutes(delta) + " minutes ago";
}
if (delta < 90L * ONE_MINUTE) {
return "one hour ago";
}
if (delta < 24L * ONE_HOUR) {
return toHours(delta) + " hours ago";
}
if (delta < 48L * ONE_HOUR) {
return "yesterday";
}
if (delta < 30L * ONE_DAY) {
return toDays(delta) + " days ago";
}
if (delta < 12L * 4L * ONE_WEEK) {
long months = toMonths(delta);
return months <= 1 ? "one month ago" : months + " months ago";
} else {
long years = toYears(delta);
return years <= 1 ? "one year ago" : years + " years ago";
}
}
private static long toSeconds(long date) {
return date / 1000L;
}
private static long toMinutes(long date) {
return toSeconds(date) / 60L;
}
private static long toHours(long date) {
return toMinutes(date) / 60L;
}
private static long toDays(long date) {
return toHours(date) / 24L;
}
private static long toMonths(long date) {
return toDays(date) / 30L;
}
private static long toYears(long date) {
return toMonths(date) / 365L;
}
}
jquery.timeago插件
Jeff,因为Stack Overflow广泛使用jQuery,所以我建议使用jQuery.timeago插件。
优点:
即使页面是在10分钟前打开的,也不要使用“1分钟前”的时间戳;timeago自动刷新。您可以充分利用web应用程序中的页面和/或片段缓存,因为时间戳不是在服务器上计算的。你可以像酷孩子一样使用微格式。
只需将其附加到DOM就绪的时间戳:
jQuery(document).ready(function() {
jQuery('abbr.timeago').timeago();
});
这将在标题中使用timeago类和ISO 8601时间戳转换所有缩写元素:
<abbr class="timeago" title="2008-07-17T09:24:17Z">July 17, 2008</abbr>
变成这样:
<abbr class="timeago" title="July 17, 2008">4 months ago</abbr>
结果:4个月前。随着时间的推移,时间戳将自动更新。
免责声明:我写了这个插件,所以我有偏见。
你可以试试这个。我想它会正常工作的。
long delta = new Date().getTime() - date.getTime();
const int SECOND = 1;
const int MINUTE = 60 * SECOND;
const int HOUR = 60 * MINUTE;
const int DAY = 24 * HOUR;
const int MONTH = 30 * DAY;
if (delta < 0L)
{
return "not yet";
}
if (delta < 1L * MINUTE)
{
return ts.Seconds == 1 ? "one second ago" : ts.Seconds + " seconds ago";
}
if (delta < 2L * MINUTE)
{
return "a minute ago";
}
if (delta < 45L * MINUTE)
{
return ts.Minutes + " minutes ago";
}
if (delta < 90L * MINUTE)
{
return "an hour ago";
}
if (delta < 24L * HOUR)
{
return ts.Hours + " hours ago";
}
if (delta < 48L * HOUR)
{
return "yesterday";
}
if (delta < 30L * DAY)
{
return ts.Days + " days ago";
}
if (delta < 12L * MONTH)
{
int months = Convert.ToInt32(Math.Floor((double)ts.Days / 30));
return months <= 1 ? "one month ago" : months + " months ago";
}
else
{
int years = Convert.ToInt32(Math.Floor((double)ts.Days / 365));
return years <= 1 ? "one year ago" : years + " years ago";
}