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
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";
}
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";
}
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()。
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);
});
};
}
我是这样做的
var ts = new TimeSpan(DateTime.UtcNow.Ticks - dt.Ticks);
double delta = Math.Abs(ts.TotalSeconds);
if (delta < 60)
{
return ts.Seconds == 1 ? "one second ago" : ts.Seconds + " seconds ago";
}
if (delta < 60 * 2)
{
return "a minute ago";
}
if (delta < 45 * 60)
{
return ts.Minutes + " minutes ago";
}
if (delta < 90 * 60)
{
return "an hour ago";
}
if (delta < 24 * 60 * 60)
{
return ts.Hours + " hours ago";
}
if (delta < 48 * 60 * 60)
{
return "yesterday";
}
if (delta < 30 * 24 * 60 * 60)
{
return ts.Days + " days ago";
}
if (delta < 12 * 30 * 24 * 60 * 60)
{
int months = Convert.ToInt32(Math.Floor((double)ts.Days / 30));
return months <= 1 ? "one month ago" : months + " months ago";
}
int years = Convert.ToInt32(Math.Floor((double)ts.Days / 365));
return years <= 1 ? "one year ago" : years + " years ago";
建议?评论?如何改进此算法?
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;
}
}