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
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;
}
其他回答
Nuget上还有一个名为Humanizr的软件包,它实际上运行得很好,并且在.NET Foundation中。
DateTime.UtcNow.AddHours(-30).Humanize() => "yesterday"
DateTime.UtcNow.AddHours(-2).Humanize() => "2 hours ago"
DateTime.UtcNow.AddHours(30).Humanize() => "tomorrow"
DateTime.UtcNow.AddHours(2).Humanize() => "2 hours from now"
TimeSpan.FromMilliseconds(1299630020).Humanize() => "2 weeks"
TimeSpan.FromMilliseconds(1299630020).Humanize(3) => "2 weeks, 1 day, 1 hour"
Scott Hanselman在他的博客上写了一篇文章
我将为此提供一些方便的扩展方法,并使代码更可读。首先,Int32的两个扩展方法。
public static class TimeSpanExtensions {
public static TimeSpan Days(this int value) {
return new TimeSpan(value, 0, 0, 0);
}
public static TimeSpan Hours(this int value) {
return new TimeSpan(0, value, 0, 0);
}
public static TimeSpan Minutes(this int value) {
return new TimeSpan(0, 0, value, 0);
}
public static TimeSpan Seconds(this int value) {
return new TimeSpan(0, 0, 0, value);
}
public static TimeSpan Milliseconds(this int value) {
return new TimeSpan(0, 0, 0, 0, value);
}
public static DateTime Ago(this TimeSpan value) {
return DateTime.Now - value;
}
}
然后,为DateTime设置一个。
public static class DateTimeExtensions {
public static DateTime Ago(this DateTime dateTime, TimeSpan delta) {
return dateTime - delta;
}
}
现在,您可以执行以下操作:
var date = DateTime.Now;
date.Ago(2.Days()); // 2 days ago
date.Ago(7.Hours()); // 7 hours ago
date.Ago(567.Milliseconds()); // 567 milliseconds ago
@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?
var ts = new TimeSpan(DateTime.Now.Ticks - dt.Ticks);
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;
}