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
下面是我作为DateTime类的扩展方法添加的一个实现,它处理未来和过去的日期,并提供了一个近似选项,允许您指定要查找的详细程度(“3小时前”与“3小时、23分钟、12秒前”):
using System.Text;
/// <summary>
/// Compares a supplied date to the current date and generates a friendly English
/// comparison ("5 days ago", "5 days from now")
/// </summary>
/// <param name="date">The date to convert</param>
/// <param name="approximate">When off, calculate timespan down to the second.
/// When on, approximate to the largest round unit of time.</param>
/// <returns></returns>
public static string ToRelativeDateString(this DateTime value, bool approximate)
{
StringBuilder sb = new StringBuilder();
string suffix = (value > DateTime.Now) ? " from now" : " ago";
TimeSpan timeSpan = new TimeSpan(Math.Abs(DateTime.Now.Subtract(value).Ticks));
if (timeSpan.Days > 0)
{
sb.AppendFormat("{0} {1}", timeSpan.Days,
(timeSpan.Days > 1) ? "days" : "day");
if (approximate) return sb.ToString() + suffix;
}
if (timeSpan.Hours > 0)
{
sb.AppendFormat("{0}{1} {2}", (sb.Length > 0) ? ", " : string.Empty,
timeSpan.Hours, (timeSpan.Hours > 1) ? "hours" : "hour");
if (approximate) return sb.ToString() + suffix;
}
if (timeSpan.Minutes > 0)
{
sb.AppendFormat("{0}{1} {2}", (sb.Length > 0) ? ", " : string.Empty,
timeSpan.Minutes, (timeSpan.Minutes > 1) ? "minutes" : "minute");
if (approximate) return sb.ToString() + suffix;
}
if (timeSpan.Seconds > 0)
{
sb.AppendFormat("{0}{1} {2}", (sb.Length > 0) ? ", " : string.Empty,
timeSpan.Seconds, (timeSpan.Seconds > 1) ? "seconds" : "second");
if (approximate) return sb.ToString() + suffix;
}
if (sb.Length == 0) return "right now";
sb.Append(suffix);
return sb.ToString();
}
其他回答
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()。
@杰夫
var ts=新时间跨度(DateTime.UtcNow.Ticks-dt.Ticks);
对DateTime执行减法仍会返回TimeSpan。
所以你可以这样做
(DateTime.UtcNow - dt).TotalSeconds
我也很惊讶地看到常数用手相乘,然后注释加上乘法。这是错误的优化吗?
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";
}
简单且100%的工作解决方案。
处理过去和将来的时间。。以防万一
public string GetTimeSince(DateTime postDate)
{
string message = "";
DateTime currentDate = DateTime.Now;
TimeSpan timegap = currentDate - postDate;
if (timegap.Days > 365)
{
message = string.Format(L("Ago") + " {0} " + L("Years"), (((timegap.Days) / 30) / 12));
}
else if (timegap.Days > 30)
{
message = string.Format(L("Ago") + " {0} " + L("Months"), timegap.Days/30);
}
else if (timegap.Days > 0)
{
message = string.Format(L("Ago") + " {0} " + L("Days"), timegap.Days);
}
else if (timegap.Hours > 0)
{
message = string.Format(L("Ago") + " {0} " + L("Hours"), timegap.Hours);
}
else if (timegap.Minutes > 0)
{
message = string.Format(L("Ago") + " {0} " + L("Minutes"), timegap.Minutes);
}
else if (timegap.Seconds > 0)
{
message = string.Format(L("Ago") + " {0} " + L("Seconds"), timegap.Seconds);
}
// let's handle future times..just in case
else if (timegap.Days < -365)
{
message = string.Format(L("In") + " {0} " + L("Years"), (((Math.Abs(timegap.Days)) / 30) / 12));
}
else if (timegap.Days < -30)
{
message = string.Format(L("In") + " {0} " + L("Months"), ((Math.Abs(timegap.Days)) / 30));
}
else if (timegap.Days < 0)
{
message = string.Format(L("In") + " {0} " + L("Days"), Math.Abs(timegap.Days));
}
else if (timegap.Hours < 0)
{
message = string.Format(L("In") + " {0} " + L("Hours"), Math.Abs(timegap.Hours));
}
else if (timegap.Minutes < 0)
{
message = string.Format(L("In") + " {0} " + L("Minutes"), Math.Abs(timegap.Minutes));
}
else if (timegap.Seconds < 0)
{
message = string.Format(L("In") + " {0} " + L("Seconds"), Math.Abs(timegap.Seconds));
}
else
{
message = "a bit";
}
return message;
}
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;
}