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
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;
}
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";
}
如果您想获得类似“2天4小时12分钟前”的输出,则需要一个时间跨度:
TimeSpan timeDiff = DateTime.Now-CreatedDate;
然后您可以访问您喜欢的值:
timeDiff.Days
timeDiff.Hours
等
@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?
您可以使用TimeAgo扩展,如下所示:
public static string TimeAgo(this DateTime dateTime)
{
string result = string.Empty;
var timeSpan = DateTime.Now.Subtract(dateTime);
if (timeSpan <= TimeSpan.FromSeconds(60))
{
result = string.Format("{0} seconds ago", timeSpan.Seconds);
}
else if (timeSpan <= TimeSpan.FromMinutes(60))
{
result = timeSpan.Minutes > 1 ?
String.Format("about {0} minutes ago", timeSpan.Minutes) :
"about a minute ago";
}
else if (timeSpan <= TimeSpan.FromHours(24))
{
result = timeSpan.Hours > 1 ?
String.Format("about {0} hours ago", timeSpan.Hours) :
"about an hour ago";
}
else if (timeSpan <= TimeSpan.FromDays(30))
{
result = timeSpan.Days > 1 ?
String.Format("about {0} days ago", timeSpan.Days) :
"yesterday";
}
else if (timeSpan <= TimeSpan.FromDays(365))
{
result = timeSpan.Days > 30 ?
String.Format("about {0} months ago", timeSpan.Days / 30) :
"about a month ago";
}
else
{
result = timeSpan.Days > 365 ?
String.Format("about {0} years ago", timeSpan.Days / 365) :
"about a year ago";
}
return result;
}
或者使用带有Timeago的Razor扩展的jQuery插件。
这里是Jeffs Script for PHP的重写:
define("SECOND", 1);
define("MINUTE", 60 * SECOND);
define("HOUR", 60 * MINUTE);
define("DAY", 24 * HOUR);
define("MONTH", 30 * DAY);
function relativeTime($time)
{
$delta = time() - $time;
if ($delta < 1 * MINUTE)
{
return $delta == 1 ? "one second ago" : $delta . " seconds ago";
}
if ($delta < 2 * MINUTE)
{
return "a minute ago";
}
if ($delta < 45 * MINUTE)
{
return floor($delta / MINUTE) . " minutes ago";
}
if ($delta < 90 * MINUTE)
{
return "an hour ago";
}
if ($delta < 24 * HOUR)
{
return floor($delta / HOUR) . " hours ago";
}
if ($delta < 48 * HOUR)
{
return "yesterday";
}
if ($delta < 30 * DAY)
{
return floor($delta / DAY) . " days ago";
}
if ($delta < 12 * MONTH)
{
$months = floor($delta / DAY / 30);
return $months <= 1 ? "one month ago" : $months . " months ago";
}
else
{
$years = floor($delta / DAY / 365);
return $years <= 1 ? "one year ago" : $years . " years ago";
}
}