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
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。
其他回答
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";
}
}
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
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
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";
}
下面是我作为DateTime类的扩展方法添加的一个实现,它处理未来和过去的日期,并提供了一个近似选项,允许您指定要查找的详细程度(“3小时前”与“3小时、23分钟、12秒前”):
using System.Text;
/// <summary>
/// Compares a supplied date to the current date and generates a friendly English
/// comparison ("5 days ago", "5 days from now")
/// </summary>
/// <param name="date">The date to convert</param>
/// <param name="approximate">When off, calculate timespan down to the second.
/// When on, approximate to the largest round unit of time.</param>
/// <returns></returns>
public static string ToRelativeDateString(this DateTime value, bool approximate)
{
StringBuilder sb = new StringBuilder();
string suffix = (value > DateTime.Now) ? " from now" : " ago";
TimeSpan timeSpan = new TimeSpan(Math.Abs(DateTime.Now.Subtract(value).Ticks));
if (timeSpan.Days > 0)
{
sb.AppendFormat("{0} {1}", timeSpan.Days,
(timeSpan.Days > 1) ? "days" : "day");
if (approximate) return sb.ToString() + suffix;
}
if (timeSpan.Hours > 0)
{
sb.AppendFormat("{0}{1} {2}", (sb.Length > 0) ? ", " : string.Empty,
timeSpan.Hours, (timeSpan.Hours > 1) ? "hours" : "hour");
if (approximate) return sb.ToString() + suffix;
}
if (timeSpan.Minutes > 0)
{
sb.AppendFormat("{0}{1} {2}", (sb.Length > 0) ? ", " : string.Empty,
timeSpan.Minutes, (timeSpan.Minutes > 1) ? "minutes" : "minute");
if (approximate) return sb.ToString() + suffix;
}
if (timeSpan.Seconds > 0)
{
sb.AppendFormat("{0}{1} {2}", (sb.Length > 0) ? ", " : string.Empty,
timeSpan.Seconds, (timeSpan.Seconds > 1) ? "seconds" : "second");
if (approximate) return sb.ToString() + suffix;
}
if (sb.Length == 0) return "right now";
sb.Append(suffix);
return sb.ToString();
}