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
通过在客户端执行此逻辑,可以减少服务器端负载。在一些Digg页面上查看源代码以供参考。它们让服务器发出一个由Javascript处理的历元时间值。这样,您就不需要管理最终用户的时区。新的服务器端代码类似于:
public string GetRelativeTime(DateTime timeStamp)
{
return string.Format("<script>printdate({0});</script>", timeStamp.ToFileTimeUtc());
}
您甚至可以在那里添加一个NOSCRIPT块,然后执行ToString()。
其他回答
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;
}
}
以某种方式,您可以使用DateTime函数以秒到年计算相对时间,请尝试以下操作:
using System;
public class Program {
public static string getRelativeTime(DateTime past) {
DateTime now = DateTime.Today;
string rt = "";
int time;
string statement = "";
if (past.Second >= now.Second) {
if (past.Second - now.Second == 1) {
rt = "second ago";
}
rt = "seconds ago";
time = past.Second - now.Second;
statement = "" + time;
return (statement + rt);
}
if (past.Minute >= now.Minute) {
if (past.Second - now.Second == 1) {
rt = "second ago";
} else {
rt = "minutes ago";
}
time = past.Minute - now.Minute;
statement = "" + time;
return (statement + rt);
}
// This process will go on until years
}
public static void Main() {
DateTime before = new DateTime(1995, 8, 24);
string date = getRelativeTime(before);
Console.WriteLine("Windows 95 was {0}.", date);
}
}
不完全有效,但如果您对其进行一点修改和调试,它很可能会完成任务。
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";
}
}
iPhone Objective-C版本
+ (NSString *)timeAgoString:(NSDate *)date {
int delta = -(int)[date timeIntervalSinceNow];
if (delta < 60)
{
return delta == 1 ? @"one second ago" : [NSString stringWithFormat:@"%i seconds ago", delta];
}
if (delta < 120)
{
return @"a minute ago";
}
if (delta < 2700)
{
return [NSString stringWithFormat:@"%i minutes ago", delta/60];
}
if (delta < 5400)
{
return @"an hour ago";
}
if (delta < 24 * 3600)
{
return [NSString stringWithFormat:@"%i hours ago", delta/3600];
}
if (delta < 48 * 3600)
{
return @"yesterday";
}
if (delta < 30 * 24 * 3600)
{
return [NSString stringWithFormat:@"%i days ago", delta/(24*3600)];
}
if (delta < 12 * 30 * 24 * 3600)
{
int months = delta/(30*24*3600);
return months <= 1 ? @"one month ago" : [NSString stringWithFormat:@"%i months ago", months];
}
else
{
int years = delta/(12*30*24*3600);
return years <= 1 ? @"one year ago" : [NSString stringWithFormat:@"%i years ago", years];
}
}
当您知道查看者的时区时,以日为单位使用日历日可能会更清晰。我不熟悉.NET库,所以我不知道如何在C#中实现这一点。
在消费者网站上,你也可以在一分钟内用手洗。“不到一分钟前”或“刚刚”就足够了。