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
使用Fluent DateTime
var dateTime1 = 2.Hours().Ago();
var dateTime2 = 3.Days().Ago();
var dateTime3 = 1.Months().Ago();
var dateTime4 = 5.Hours().FromNow();
var dateTime5 = 2.Weeks().FromNow();
var dateTime6 = 40.Seconds().FromNow();
其他回答
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?
@杰夫
我知道你的有点长。然而,随着对“昨天”和“几年”的支持,它似乎确实更为有力。但根据我的经验,当使用此选项时,用户最有可能在前30天内查看内容。只有真正的铁杆人才会在这之后出现。所以,我通常选择保持简短。
这是我目前在我的一个网站上使用的方法。这只返回相对的日期、小时和时间。然后用户必须在输出中加上“ago”。
public static string ToLongString(this TimeSpan time)
{
string output = String.Empty;
if (time.Days > 0)
output += time.Days + " days ";
if ((time.Days == 0 || time.Days == 1) && time.Hours > 0)
output += time.Hours + " hr ";
if (time.Days == 0 && time.Minutes > 0)
output += time.Minutes + " min ";
if (output.Length == 0)
output += time.Seconds + " sec";
return output.Trim();
}
这是我的功能,就像一个魅力:)
public static string RelativeDate(DateTime theDate)
{
var span = DateTime.Now - theDate;
if (span.Days > 365)
{
var years = (span.Days / 365);
if (span.Days % 365 != 0)
years += 1;
return $"about {years} {(years == 1 ? "year" : "years")} ago";
}
if (span.Days > 30)
{
var months = (span.Days / 30);
if (span.Days % 31 != 0)
months += 1;
return $"about {months} {(months == 1 ? "month" : "months")} ago";
}
if (span.Days > 0)
return $"about {span.Days} {(span.Days == 1 ? "day" : "days")} ago";
if (span.Hours > 0)
return $"about {span.Hours} {(span.Hours == 1 ? "hour" : "hours")} ago";
if (span.Minutes > 0)
return $"about {span.Minutes} {(span.Minutes == 1 ? "minute" : "minutes")} ago";
if (span.Seconds > 5)
return $"about {span.Seconds} seconds ago";
return span.Seconds <= 5 ? "about 5 seconds ago" : string.Empty;
}
/**
* {@code date1} has to be earlier than {@code date2}.
*/
public static String relativize(Date date1, Date date2) {
assert date2.getTime() >= date1.getTime();
long duration = date2.getTime() - date1.getTime();
long converted;
if ((converted = TimeUnit.MILLISECONDS.toDays(duration)) > 0) {
return String.format("%d %s ago", converted, converted == 1 ? "day" : "days");
} else if ((converted = TimeUnit.MILLISECONDS.toHours(duration)) > 0) {
return String.format("%d %s ago", converted, converted == 1 ? "hour" : "hours");
} else if ((converted = TimeUnit.MILLISECONDS.toMinutes(duration)) > 0) {
return String.format("%d %s ago", converted, converted == 1 ? "minute" : "minutes");
} else if ((converted = TimeUnit.MILLISECONDS.toSeconds(duration)) > 0) {
return String.format("%d %s ago", converted, converted == 1 ? "second" : "seconds");
} else {
return "just now";
}
}
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;
}