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
var ts = new TimeSpan(DateTime.Now.Ticks - dt.Ticks);
其他回答
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)
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;
}
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()。
聚会晚了几年,但我有一个要求,无论是过去还是将来的约会,都要这样做,所以我把杰夫和文森特的约会结合在一起。这是一场盛大的盛会!:)
public static class DateTimeHelper
{
private const int SECOND = 1;
private const int MINUTE = 60 * SECOND;
private const int HOUR = 60 * MINUTE;
private const int DAY = 24 * HOUR;
private const int MONTH = 30 * DAY;
/// <summary>
/// Returns a friendly version of the provided DateTime, relative to now. E.g.: "2 days ago", or "in 6 months".
/// </summary>
/// <param name="dateTime">The DateTime to compare to Now</param>
/// <returns>A friendly string</returns>
public static string GetFriendlyRelativeTime(DateTime dateTime)
{
if (DateTime.UtcNow.Ticks == dateTime.Ticks)
{
return "Right now!";
}
bool isFuture = (DateTime.UtcNow.Ticks < dateTime.Ticks);
var ts = DateTime.UtcNow.Ticks < dateTime.Ticks ? new TimeSpan(dateTime.Ticks - DateTime.UtcNow.Ticks) : new TimeSpan(DateTime.UtcNow.Ticks - dateTime.Ticks);
double delta = ts.TotalSeconds;
if (delta < 1 * MINUTE)
{
return isFuture ? "in " + (ts.Seconds == 1 ? "one second" : ts.Seconds + " seconds") : ts.Seconds == 1 ? "one second ago" : ts.Seconds + " seconds ago";
}
if (delta < 2 * MINUTE)
{
return isFuture ? "in a minute" : "a minute ago";
}
if (delta < 45 * MINUTE)
{
return isFuture ? "in " + ts.Minutes + " minutes" : ts.Minutes + " minutes ago";
}
if (delta < 90 * MINUTE)
{
return isFuture ? "in an hour" : "an hour ago";
}
if (delta < 24 * HOUR)
{
return isFuture ? "in " + ts.Hours + " hours" : ts.Hours + " hours ago";
}
if (delta < 48 * HOUR)
{
return isFuture ? "tomorrow" : "yesterday";
}
if (delta < 30 * DAY)
{
return isFuture ? "in " + ts.Days + " days" : ts.Days + " days ago";
}
if (delta < 12 * MONTH)
{
int months = Convert.ToInt32(Math.Floor((double)ts.Days / 30));
return isFuture ? "in " + (months <= 1 ? "one month" : months + " months") : months <= 1 ? "one month ago" : months + " months ago";
}
else
{
int years = Convert.ToInt32(Math.Floor((double)ts.Days / 365));
return isFuture ? "in " + (years <= 1 ? "one year" : years + " years") : years <= 1 ? "one year ago" : years + " years ago";
}
}
}
这是stackoverflow使用的算法,但使用了错误修复(没有“一小时前”)的perlish伪代码进行了更简洁的重写。该函数在秒前取一个(正数),并返回一个人类友好的字符串,如“3小时前”或“昨天”。
agoify($delta)
local($y, $mo, $d, $h, $m, $s);
$s = floor($delta);
if($s<=1) return "a second ago";
if($s<60) return "$s seconds ago";
$m = floor($s/60);
if($m==1) return "a minute ago";
if($m<45) return "$m minutes ago";
$h = floor($m/60);
if($h==1) return "an hour ago";
if($h<24) return "$h hours ago";
$d = floor($h/24);
if($d<2) return "yesterday";
if($d<30) return "$d days ago";
$mo = floor($d/30);
if($mo<=1) return "a month ago";
$y = floor($mo/12);
if($y<1) return "$mo months ago";
if($y==1) return "a year ago";
return "$y years ago";