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;
}
其他回答
Jeff,您的代码很好,但使用常量可以更清晰(如代码完成中所建议的)。
const int SECOND = 1;
const int MINUTE = 60 * SECOND;
const int HOUR = 60 * MINUTE;
const int DAY = 24 * HOUR;
const int MONTH = 30 * DAY;
var ts = new TimeSpan(DateTime.UtcNow.Ticks - yourDate.Ticks);
double delta = Math.Abs(ts.TotalSeconds);
if (delta < 1 * MINUTE)
return ts.Seconds == 1 ? "one second ago" : ts.Seconds + " seconds ago";
if (delta < 2 * MINUTE)
return "a minute ago";
if (delta < 45 * MINUTE)
return ts.Minutes + " minutes ago";
if (delta < 90 * MINUTE)
return "an hour ago";
if (delta < 24 * HOUR)
return ts.Hours + " hours ago";
if (delta < 48 * HOUR)
return "yesterday";
if (delta < 30 * DAY)
return ts.Days + " days ago";
if (delta < 12 * 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";
}
鉴于全世界和她的丈夫似乎都在发布代码样本,这是我不久前根据这些答案写的。
我特别需要这个代码可以本地化。所以我有两个类——Grammar,它指定了可本地化的术语,FuzzyDateExtensions,它包含一系列扩展方法。我不需要处理未来的日期时间,因此不尝试使用此代码处理它们。
为了简洁起见,我在源代码中保留了一些XMLdoc,但删除了大部分(显而易见的地方)。我也没有把每个班级成员都包括在这里:
public class Grammar
{
/// <summary> Gets or sets the term for "just now". </summary>
public string JustNow { get; set; }
/// <summary> Gets or sets the term for "X minutes ago". </summary>
/// <remarks>
/// This is a <see cref="String.Format"/> pattern, where <c>{0}</c>
/// is the number of minutes.
/// </remarks>
public string MinutesAgo { get; set; }
public string OneHourAgo { get; set; }
public string HoursAgo { get; set; }
public string Yesterday { get; set; }
public string DaysAgo { get; set; }
public string LastMonth { get; set; }
public string MonthsAgo { get; set; }
public string LastYear { get; set; }
public string YearsAgo { get; set; }
/// <summary> Gets or sets the term for "ages ago". </summary>
public string AgesAgo { get; set; }
/// <summary>
/// Gets or sets the threshold beyond which the fuzzy date should be
/// considered "ages ago".
/// </summary>
public TimeSpan AgesAgoThreshold { get; set; }
/// <summary>
/// Initialises a new <see cref="Grammar"/> instance with the
/// specified properties.
/// </summary>
private void Initialise(string justNow, string minutesAgo,
string oneHourAgo, string hoursAgo, string yesterday, string daysAgo,
string lastMonth, string monthsAgo, string lastYear, string yearsAgo,
string agesAgo, TimeSpan agesAgoThreshold)
{ ... }
}
FuzzyDateString类包含:
public static class FuzzyDateExtensions
{
public static string ToFuzzyDateString(this TimeSpan timespan)
{
return timespan.ToFuzzyDateString(new Grammar());
}
public static string ToFuzzyDateString(this TimeSpan timespan,
Grammar grammar)
{
return GetFuzzyDateString(timespan, grammar);
}
public static string ToFuzzyDateString(this DateTime datetime)
{
return (DateTime.Now - datetime).ToFuzzyDateString();
}
public static string ToFuzzyDateString(this DateTime datetime,
Grammar grammar)
{
return (DateTime.Now - datetime).ToFuzzyDateString(grammar);
}
private static string GetFuzzyDateString(TimeSpan timespan,
Grammar grammar)
{
timespan = timespan.Duration();
if (timespan >= grammar.AgesAgoThreshold)
{
return grammar.AgesAgo;
}
if (timespan < new TimeSpan(0, 2, 0)) // 2 minutes
{
return grammar.JustNow;
}
if (timespan < new TimeSpan(1, 0, 0)) // 1 hour
{
return String.Format(grammar.MinutesAgo, timespan.Minutes);
}
if (timespan < new TimeSpan(1, 55, 0)) // 1 hour 55 minutes
{
return grammar.OneHourAgo;
}
if (timespan < new TimeSpan(12, 0, 0) // 12 hours
&& (DateTime.Now - timespan).IsToday())
{
return String.Format(grammar.HoursAgo, timespan.RoundedHours());
}
if ((DateTime.Now.AddDays(1) - timespan).IsToday())
{
return grammar.Yesterday;
}
if (timespan < new TimeSpan(32, 0, 0, 0) // 32 days
&& (DateTime.Now - timespan).IsThisMonth())
{
return String.Format(grammar.DaysAgo, timespan.RoundedDays());
}
if ((DateTime.Now.AddMonths(1) - timespan).IsThisMonth())
{
return grammar.LastMonth;
}
if (timespan < new TimeSpan(365, 0, 0, 0, 0) // 365 days
&& (DateTime.Now - timespan).IsThisYear())
{
return String.Format(grammar.MonthsAgo, timespan.RoundedMonths());
}
if ((DateTime.Now - timespan).AddYears(1).IsThisYear())
{
return grammar.LastYear;
}
return String.Format(grammar.YearsAgo, timespan.RoundedYears());
}
}
我想实现的一件关键事情,以及本地化,就是“今天”只意味着“这个日历日”,所以IsToday、IsThisMonth和IsThisYear方法如下:
public static bool IsToday(this DateTime date)
{
return date.DayOfYear == DateTime.Now.DayOfYear && date.IsThisYear();
}
舍入方法如下(我已包括RoundedMonths,因为这有点不同):
public static int RoundedDays(this TimeSpan timespan)
{
return (timespan.Hours > 12) ? timespan.Days + 1 : timespan.Days;
}
public static int RoundedMonths(this TimeSpan timespan)
{
DateTime then = DateTime.Now - timespan;
// Number of partial months elapsed since 1 Jan, AD 1 (DateTime.MinValue)
int nowMonthYears = DateTime.Now.Year * 12 + DateTime.Now.Month;
int thenMonthYears = then.Year * 12 + then.Month;
return nowMonthYears - thenMonthYears;
}
我希望人们觉得这很有用和/或有趣:o)
如果您想获得类似“2天4小时12分钟前”的输出,则需要一个时间跨度:
TimeSpan timeDiff = DateTime.Now-CreatedDate;
然后您可以访问您喜欢的值:
timeDiff.Days
timeDiff.Hours
等
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。
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";
} ?>