使用解构主义和Linq得到“n（最大时间单位）前”的“一行”：

TimeSpan timeSpan = DateTime.Now - new DateTime(1234, 5, 6, 7, 8, 9);

(string unit, int value) = new Dictionary<string, int>
{
    {"year(s)", (int)(timeSpan.TotalDays / 365.25)}, //https://en.wikipedia.org/wiki/Year#Intercalation
    {"month(s)", (int)(timeSpan.TotalDays / 29.53)}, //https://en.wikipedia.org/wiki/Month
    {"day(s)", (int)timeSpan.TotalDays},
    {"hour(s)", (int)timeSpan.TotalHours},
    {"minute(s)", (int)timeSpan.TotalMinutes},
    {"second(s)", (int)timeSpan.TotalSeconds},
    {"millisecond(s)", (int)timeSpan.TotalMilliseconds}
}.First(kvp => kvp.Value > 0);

Console.WriteLine($"{value} {unit} ago");

你在786年前

当前年份和月份，如

TimeSpan timeSpan = DateTime.Now - new DateTime(2020, 12, 6, 7, 8, 9);

您4天前收到

实际日期，比如

TimeSpan timeSpan = DateTime.Now - DateTime.Now.Date;

9小时前到达

文森特接受的答案做出了许多武断的决定。为什么45分钟舍入为一小时，而45秒不舍入为一分钟？在年和月的计算中，它的圈复杂度增加了，这使得遵循逻辑变得更加复杂。它假设TimeSpan是相对于过去（2天前）的，而它很可能是在未来（2天后）。它定义了不必要的常量，而不是使用TimeSpan.TicksPerSecond等。

此实现解决了上述问题，并更新了语法以使用开关表达式和关系模式

/// <summary>
/// Convert a <see cref="TimeSpan"/> to a natural language representation.
/// </summary>
/// <example>
/// <code>
/// TimeSpan.FromSeconds(10).ToNaturalLanguage();
/// // 10 seconds
/// </code>
/// </example>
public static string ToNaturalLanguage(this TimeSpan @this)
{
    const int daysInWeek = 7;
    const int daysInMonth = 30;
    const int daysInYear = 365;
    const long threshold = 100 * TimeSpan.TicksPerMillisecond;
    @this = @this.TotalSeconds < 0
        ? TimeSpan.FromSeconds(@this.TotalSeconds * -1)
        : @this;
    return (@this.Ticks + threshold) switch
    {
        < 2 * TimeSpan.TicksPerSecond => "a second",
        < 1 * TimeSpan.TicksPerMinute => @this.Seconds + " seconds",
        < 2 * TimeSpan.TicksPerMinute => "a minute",
        < 1 * TimeSpan.TicksPerHour => @this.Minutes + " minutes",
        < 2 * TimeSpan.TicksPerHour => "an hour",
        < 1 * TimeSpan.TicksPerDay => @this.Hours + " hours",
        < 2 * TimeSpan.TicksPerDay => "a day",
        < 1 * daysInWeek * TimeSpan.TicksPerDay => @this.Days + " days",
        < 2 * daysInWeek * TimeSpan.TicksPerDay => "a week",
        < 1 * daysInMonth * TimeSpan.TicksPerDay => (@this.Days / daysInWeek).ToString("F0") + " weeks",
        < 2 * daysInMonth * TimeSpan.TicksPerDay => "a month",
        < 1 * daysInYear * TimeSpan.TicksPerDay => (@this.Days / daysInMonth).ToString("F0") + " months",
        < 2 * daysInYear * TimeSpan.TicksPerDay => "a year",
        _ => (@this.Days / daysInYear).ToString("F0") + " years"
    };
}

/// <summary>
/// Convert a <see cref="DateTime"/> to a natural language representation.
/// </summary>
/// <example>
/// <code>
/// (DateTime.Now - TimeSpan.FromSeconds(10)).ToNaturalLanguage()
/// // 10 seconds ago
/// </code>
/// </example>
public static string ToNaturalLanguage(this DateTime @this)
{
    TimeSpan timeSpan = @this - DateTime.Now;
    return timeSpan.TotalSeconds switch
    {
        >= 1 => timeSpan.ToNaturalLanguage() + " until",
        <= -1 => timeSpan.ToNaturalLanguage() + " ago",
        _ => "now",
    };
}

可以使用NUnit对其进行如下测试：

[TestCase("a second", 0)]
[TestCase("a second", 1)]
[TestCase("2 seconds", 2)]
[TestCase("a minute", 0, 1)]
[TestCase("5 minutes", 0, 5)]
[TestCase("an hour", 0, 0, 1)]
[TestCase("2 hours", 0, 0, 2)]
[TestCase("a day", 0, 0, 24)]
[TestCase("a day", 0, 0, 0, 1)]
[TestCase("6 days", 0, 0, 0, 6)]
[TestCase("a week", 0, 0, 0, 7)]
[TestCase("4 weeks", 0, 0, 0, 29)]
[TestCase("a month", 0, 0, 0, 30)]
[TestCase("6 months", 0, 0, 0, 6 * 30)]
[TestCase("a year", 0, 0, 0, 365)]
[TestCase("68 years", int.MaxValue)]
public void NaturalLanguageHelpers_TimeSpan(
    string expected,
    int seconds,
    int minutes = 0,
    int hours = 0,
    int days = 0
)
{
    // Arrange
    TimeSpan timeSpan = new(days, hours, minutes, seconds);

    // Act
    string result = timeSpan.ToNaturalLanguage();

    // Assert
    Assert.That(result, Is.EqualTo(expected));
}

[TestCase("now", 0)]
[TestCase("10 minutes ago", 0, -10)]
[TestCase("10 minutes until", 10, 10)]
[TestCase("68 years until", int.MaxValue)]
[TestCase("68 years ago", int.MinValue)]
public void NaturalLanguageHelpers_DateTime(
    string expected,
    int seconds,
    int minutes = 0,
    int hours = 0,
    int days = 0
)
{
    // Arrange
    TimeSpan timeSpan = new(days, hours, minutes, seconds);
    DateTime now = DateTime.Now;
    DateTime dateTime = now + timeSpan;

    // Act
    string result = dateTime.ToNaturalLanguage();

    // Assert
    Assert.That(result, Is.EqualTo(expected));
}

或者作为要点：https://gist.github.com/StudioLE/2dd394e3f792e79adc927ede274df56e

@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？

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";
    }
}    

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;
}