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
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。
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
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。
这是我的功能,就像一个魅力:)
public static string RelativeDate(DateTime theDate)
{
var span = DateTime.Now - theDate;
if (span.Days > 365)
{
var years = (span.Days / 365);
if (span.Days % 365 != 0)
years += 1;
return $"about {years} {(years == 1 ? "year" : "years")} ago";
}
if (span.Days > 30)
{
var months = (span.Days / 30);
if (span.Days % 31 != 0)
months += 1;
return $"about {months} {(months == 1 ? "month" : "months")} ago";
}
if (span.Days > 0)
return $"about {span.Days} {(span.Days == 1 ? "day" : "days")} ago";
if (span.Hours > 0)
return $"about {span.Hours} {(span.Hours == 1 ? "hour" : "hours")} ago";
if (span.Minutes > 0)
return $"about {span.Minutes} {(span.Minutes == 1 ? "minute" : "minutes")} ago";
if (span.Seconds > 5)
return $"about {span.Seconds} seconds ago";
return span.Seconds <= 5 ? "about 5 seconds ago" : string.Empty;
}