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
简单且100%的工作解决方案。
处理过去和将来的时间。。以防万一
public string GetTimeSince(DateTime postDate)
{
string message = "";
DateTime currentDate = DateTime.Now;
TimeSpan timegap = currentDate - postDate;
if (timegap.Days > 365)
{
message = string.Format(L("Ago") + " {0} " + L("Years"), (((timegap.Days) / 30) / 12));
}
else if (timegap.Days > 30)
{
message = string.Format(L("Ago") + " {0} " + L("Months"), timegap.Days/30);
}
else if (timegap.Days > 0)
{
message = string.Format(L("Ago") + " {0} " + L("Days"), timegap.Days);
}
else if (timegap.Hours > 0)
{
message = string.Format(L("Ago") + " {0} " + L("Hours"), timegap.Hours);
}
else if (timegap.Minutes > 0)
{
message = string.Format(L("Ago") + " {0} " + L("Minutes"), timegap.Minutes);
}
else if (timegap.Seconds > 0)
{
message = string.Format(L("Ago") + " {0} " + L("Seconds"), timegap.Seconds);
}
// let's handle future times..just in case
else if (timegap.Days < -365)
{
message = string.Format(L("In") + " {0} " + L("Years"), (((Math.Abs(timegap.Days)) / 30) / 12));
}
else if (timegap.Days < -30)
{
message = string.Format(L("In") + " {0} " + L("Months"), ((Math.Abs(timegap.Days)) / 30));
}
else if (timegap.Days < 0)
{
message = string.Format(L("In") + " {0} " + L("Days"), Math.Abs(timegap.Days));
}
else if (timegap.Hours < 0)
{
message = string.Format(L("In") + " {0} " + L("Hours"), Math.Abs(timegap.Hours));
}
else if (timegap.Minutes < 0)
{
message = string.Format(L("In") + " {0} " + L("Minutes"), Math.Abs(timegap.Minutes));
}
else if (timegap.Seconds < 0)
{
message = string.Format(L("In") + " {0} " + L("Seconds"), Math.Abs(timegap.Seconds));
}
else
{
message = "a bit";
}
return message;
}
其他回答
聚会晚了几年,但我有一个要求,无论是过去还是将来的约会,都要这样做,所以我把杰夫和文森特的约会结合在一起。这是一场盛大的盛会!:)
public static class DateTimeHelper
{
private const int SECOND = 1;
private const int MINUTE = 60 * SECOND;
private const int HOUR = 60 * MINUTE;
private const int DAY = 24 * HOUR;
private const int MONTH = 30 * DAY;
/// <summary>
/// Returns a friendly version of the provided DateTime, relative to now. E.g.: "2 days ago", or "in 6 months".
/// </summary>
/// <param name="dateTime">The DateTime to compare to Now</param>
/// <returns>A friendly string</returns>
public static string GetFriendlyRelativeTime(DateTime dateTime)
{
if (DateTime.UtcNow.Ticks == dateTime.Ticks)
{
return "Right now!";
}
bool isFuture = (DateTime.UtcNow.Ticks < dateTime.Ticks);
var ts = DateTime.UtcNow.Ticks < dateTime.Ticks ? new TimeSpan(dateTime.Ticks - DateTime.UtcNow.Ticks) : new TimeSpan(DateTime.UtcNow.Ticks - dateTime.Ticks);
double delta = ts.TotalSeconds;
if (delta < 1 * MINUTE)
{
return isFuture ? "in " + (ts.Seconds == 1 ? "one second" : ts.Seconds + " seconds") : ts.Seconds == 1 ? "one second ago" : ts.Seconds + " seconds ago";
}
if (delta < 2 * MINUTE)
{
return isFuture ? "in a minute" : "a minute ago";
}
if (delta < 45 * MINUTE)
{
return isFuture ? "in " + ts.Minutes + " minutes" : ts.Minutes + " minutes ago";
}
if (delta < 90 * MINUTE)
{
return isFuture ? "in an hour" : "an hour ago";
}
if (delta < 24 * HOUR)
{
return isFuture ? "in " + ts.Hours + " hours" : ts.Hours + " hours ago";
}
if (delta < 48 * HOUR)
{
return isFuture ? "tomorrow" : "yesterday";
}
if (delta < 30 * DAY)
{
return isFuture ? "in " + ts.Days + " days" : ts.Days + " days ago";
}
if (delta < 12 * MONTH)
{
int months = Convert.ToInt32(Math.Floor((double)ts.Days / 30));
return isFuture ? "in " + (months <= 1 ? "one month" : months + " months") : months <= 1 ? "one month ago" : months + " months ago";
}
else
{
int years = Convert.ToInt32(Math.Floor((double)ts.Days / 365));
return isFuture ? "in " + (years <= 1 ? "one year" : years + " years") : years <= 1 ? "one year ago" : years + " years ago";
}
}
}
通过在客户端执行此逻辑,可以减少服务器端负载。在一些Digg页面上查看源代码以供参考。它们让服务器发出一个由Javascript处理的历元时间值。这样,您就不需要管理最终用户的时区。新的服务器端代码类似于:
public string GetRelativeTime(DateTime timeStamp)
{
return string.Format("<script>printdate({0});</script>", timeStamp.ToFileTimeUtc());
}
您甚至可以在那里添加一个NOSCRIPT块,然后执行ToString()。
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
iPhone Objective-C版本
+ (NSString *)timeAgoString:(NSDate *)date {
int delta = -(int)[date timeIntervalSinceNow];
if (delta < 60)
{
return delta == 1 ? @"one second ago" : [NSString stringWithFormat:@"%i seconds ago", delta];
}
if (delta < 120)
{
return @"a minute ago";
}
if (delta < 2700)
{
return [NSString stringWithFormat:@"%i minutes ago", delta/60];
}
if (delta < 5400)
{
return @"an hour ago";
}
if (delta < 24 * 3600)
{
return [NSString stringWithFormat:@"%i hours ago", delta/3600];
}
if (delta < 48 * 3600)
{
return @"yesterday";
}
if (delta < 30 * 24 * 3600)
{
return [NSString stringWithFormat:@"%i days ago", delta/(24*3600)];
}
if (delta < 12 * 30 * 24 * 3600)
{
int months = delta/(30*24*3600);
return months <= 1 ? @"one month ago" : [NSString stringWithFormat:@"%i months ago", months];
}
else
{
int years = delta/(12*30*24*3600);
return years <= 1 ? @"one year ago" : [NSString stringWithFormat:@"%i years ago", years];
}
}