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

/** 
 * {@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";
    }
}

其他回答

当然,解决“1小时前”问题的一个简单方法是增加“一小时前”有效的窗口。改变

if (delta < 5400) // 90 * 60
{
    return "an hour ago";
}

into

if (delta < 7200) // 120 * 60
{
    return "an hour ago";
}

这意味着110分钟前发生的事情将被解读为“一小时前”——这可能并不完美,但我认为这比“1小时前”的现状要好。

@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?

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();
public static string ToRelativeDate(DateTime input)
{
    TimeSpan oSpan = DateTime.Now.Subtract(input);
    double TotalMinutes = oSpan.TotalMinutes;
    string Suffix = " ago";

    if (TotalMinutes < 0.0)
    {
        TotalMinutes = Math.Abs(TotalMinutes);
        Suffix = " from now";
    }

    var aValue = new SortedList<double, Func<string>>();
    aValue.Add(0.75, () => "less than a minute");
    aValue.Add(1.5, () => "about a minute");
    aValue.Add(45, () => string.Format("{0} minutes", Math.Round(TotalMinutes)));
    aValue.Add(90, () => "about an hour");
    aValue.Add(1440, () => string.Format("about {0} hours", Math.Round(Math.Abs(oSpan.TotalHours)))); // 60 * 24
    aValue.Add(2880, () => "a day"); // 60 * 48
    aValue.Add(43200, () => string.Format("{0} days", Math.Floor(Math.Abs(oSpan.TotalDays)))); // 60 * 24 * 30
    aValue.Add(86400, () => "about a month"); // 60 * 24 * 60
    aValue.Add(525600, () => string.Format("{0} months", Math.Floor(Math.Abs(oSpan.TotalDays / 30)))); // 60 * 24 * 365 
    aValue.Add(1051200, () => "about a year"); // 60 * 24 * 365 * 2
    aValue.Add(double.MaxValue, () => string.Format("{0} years", Math.Floor(Math.Abs(oSpan.TotalDays / 365))));

    return aValue.First(n => TotalMinutes < n.Key).Value.Invoke() + Suffix;
}

http://refactormycode.com/codes/493-twitter-esque-relative-dates

C#6版本:

static readonly SortedList<double, Func<TimeSpan, string>> offsets = 
   new SortedList<double, Func<TimeSpan, string>>
{
    { 0.75, _ => "less than a minute"},
    { 1.5, _ => "about a minute"},
    { 45, x => $"{x.TotalMinutes:F0} minutes"},
    { 90, x => "about an hour"},
    { 1440, x => $"about {x.TotalHours:F0} hours"},
    { 2880, x => "a day"},
    { 43200, x => $"{x.TotalDays:F0} days"},
    { 86400, x => "about a month"},
    { 525600, x => $"{x.TotalDays / 30:F0} months"},
    { 1051200, x => "about a year"},
    { double.MaxValue, x => $"{x.TotalDays / 365:F0} years"}
};

public static string ToRelativeDate(this DateTime input)
{
    TimeSpan x = DateTime.Now - input;
    string Suffix = x.TotalMinutes > 0 ? " ago" : " from now";
    x = new TimeSpan(Math.Abs(x.Ticks));
    return offsets.First(n => x.TotalMinutes < n.Key).Value(x) + Suffix;
}

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;
}