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
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;
}
}
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";
}
}
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;
}
}
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
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。
在Java中有没有一种简单的方法可以做到这一点？java.util.Date类似乎相当有限。
下面是我的快速而肮脏的Java解决方案：
import java.util.Date;
import javax.management.timer.Timer;
String getRelativeDate(Date date) {
long delta = new Date().getTime() - date.getTime();
if (delta < 1L * Timer.ONE_MINUTE) {
return toSeconds(delta) == 1 ? "one second ago" : toSeconds(delta) + " seconds ago";
}
if (delta < 2L * Timer.ONE_MINUTE) {
return "a minute ago";
}
if (delta < 45L * Timer.ONE_MINUTE) {
return toMinutes(delta) + " minutes ago";
}
if (delta < 90L * Timer.ONE_MINUTE) {
return "an hour ago";
}
if (delta < 24L * Timer.ONE_HOUR) {
return toHours(delta) + " hours ago";
}
if (delta < 48L * Timer.ONE_HOUR) {
return "yesterday";
}
if (delta < 30L * Timer.ONE_DAY) {
return toDays(delta) + " days ago";
}
if (delta < 12L * 4L * Timer.ONE_WEEK) { // a month
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 long toSeconds(long date) {
return date / 1000L;
}
private long toMinutes(long date) {
return toSeconds(date) / 60L;
}
private long toHours(long date) {
return toMinutes(date) / 60L;
}
private long toDays(long date) {
return toHours(date) / 24L;
}
private long toMonths(long date) {
return toDays(date) / 30L;
}
private long toYears(long date) {
return toMonths(date) / 365L;
}
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