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

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

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?

这是stackoverflow使用的算法,但使用了错误修复(没有“一小时前”)的perlish伪代码进行了更简洁的重写。该函数在秒前取一个(正数),并返回一个人类友好的字符串,如“3小时前”或“昨天”。

agoify($delta)
  local($y, $mo, $d, $h, $m, $s);
  $s = floor($delta);
  if($s<=1)            return "a second ago";
  if($s<60)            return "$s seconds ago";
  $m = floor($s/60);
  if($m==1)            return "a minute ago";
  if($m<45)            return "$m minutes ago";
  $h = floor($m/60);
  if($h==1)            return "an hour ago";
  if($h<24)            return "$h hours ago";
  $d = floor($h/24);
  if($d<2)             return "yesterday";
  if($d<30)            return "$d days ago";
  $mo = floor($d/30);
  if($mo<=1)           return "a month ago";
  $y = floor($mo/12);
  if($y<1)             return "$mo months ago";
  if($y==1)            return "a year ago";
  return "$y years ago";

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

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();
using System;
using System.Collections.Generic;
using System.Linq;

public static class RelativeDateHelper
{
    private static Dictionary<double, Func<double, string>> sm_Dict = null;

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

        return dict;
    }

    public static string ToRelativeDate(this 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";
        }

        if (null == sm_Dict)
            sm_Dict = DictionarySetup();

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

与此问题的另一个答案相同,但作为静态字典的扩展方法。