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
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
其他回答
你可以试试这个。我想它会正常工作的。
long delta = new Date().getTime() - date.getTime();
const int SECOND = 1;
const int MINUTE = 60 * SECOND;
const int HOUR = 60 * MINUTE;
const int DAY = 24 * HOUR;
const int MONTH = 30 * DAY;
if (delta < 0L)
{
return "not yet";
}
if (delta < 1L * MINUTE)
{
return ts.Seconds == 1 ? "one second ago" : ts.Seconds + " seconds ago";
}
if (delta < 2L * MINUTE)
{
return "a minute ago";
}
if (delta < 45L * MINUTE)
{
return ts.Minutes + " minutes ago";
}
if (delta < 90L * MINUTE)
{
return "an hour ago";
}
if (delta < 24L * HOUR)
{
return ts.Hours + " hours ago";
}
if (delta < 48L * HOUR)
{
return "yesterday";
}
if (delta < 30L * DAY)
{
return ts.Days + " days ago";
}
if (delta < 12L * 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";
}
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;
}
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。
我的方法要简单得多。您可以根据需要调整返回字符串
public static string TimeLeft(DateTime utcDate)
{
TimeSpan timeLeft = DateTime.UtcNow - utcDate;
string timeLeftString = "";
if (timeLeft.Days > 0)
{
timeLeftString += timeLeft.Days == 1 ? timeLeft.Days + " day" : timeLeft.Days + " days";
}
else if (timeLeft.Hours > 0)
{
timeLeftString += timeLeft.Hours == 1 ? timeLeft.Hours + " hour" : timeLeft.Hours + " hours";
}
else
{
timeLeftString += timeLeft.Minutes == 1 ? timeLeft.Minutes+" minute" : timeLeft.Minutes + " minutes";
}
return timeLeftString;
}
我也建议在客户端进行计算。服务器工作更少。
以下是我使用的版本(来自Zach Leatherman)
/*
* Javascript Humane Dates
* Copyright (c) 2008 Dean Landolt (deanlandolt.com)
* Re-write by Zach Leatherman (zachleat.com)
*
* Adopted from the John Resig's pretty.js
* at http://ejohn.org/blog/javascript-pretty-date
* and henrah's proposed modification
* at http://ejohn.org/blog/javascript-pretty-date/#comment-297458
*
* Licensed under the MIT license.
*/
function humane_date(date_str){
var time_formats = [
[60, 'just now'],
[90, '1 minute'], // 60*1.5
[3600, 'minutes', 60], // 60*60, 60
[5400, '1 hour'], // 60*60*1.5
[86400, 'hours', 3600], // 60*60*24, 60*60
[129600, '1 day'], // 60*60*24*1.5
[604800, 'days', 86400], // 60*60*24*7, 60*60*24
[907200, '1 week'], // 60*60*24*7*1.5
[2628000, 'weeks', 604800], // 60*60*24*(365/12), 60*60*24*7
[3942000, '1 month'], // 60*60*24*(365/12)*1.5
[31536000, 'months', 2628000], // 60*60*24*365, 60*60*24*(365/12)
[47304000, '1 year'], // 60*60*24*365*1.5
[3153600000, 'years', 31536000], // 60*60*24*365*100, 60*60*24*365
[4730400000, '1 century'] // 60*60*24*365*100*1.5
];
var time = ('' + date_str).replace(/-/g,"/").replace(/[TZ]/g," "),
dt = new Date,
seconds = ((dt - new Date(time) + (dt.getTimezoneOffset() * 60000)) / 1000),
token = ' ago',
i = 0,
format;
if (seconds < 0) {
seconds = Math.abs(seconds);
token = '';
}
while (format = time_formats[i++]) {
if (seconds < format[0]) {
if (format.length == 2) {
return format[1] + (i > 1 ? token : ''); // Conditional so we don't return Just Now Ago
} else {
return Math.round(seconds / format[2]) + ' ' + format[1] + (i > 1 ? token : '');
}
}
}
// overflow for centuries
if(seconds > 4730400000)
return Math.round(seconds / 4730400000) + ' centuries' + token;
return date_str;
};
if(typeof jQuery != 'undefined') {
jQuery.fn.humane_dates = function(){
return this.each(function(){
var date = humane_date(this.title);
if(date && jQuery(this).text() != date) // don't modify the dom if we don't have to
jQuery(this).text(date);
});
};
}