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

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

其他回答

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;
}

这是我的功能,就像一个魅力:)

public static string RelativeDate(DateTime theDate)
{
   var span = DateTime.Now - theDate;
   if (span.Days > 365)
   {
      var years = (span.Days / 365);
      if (span.Days % 365 != 0)
         years += 1;
      return $"about {years} {(years == 1 ? "year" : "years")} ago";
   }
   if (span.Days > 30)
   {
      var months = (span.Days / 30);
      if (span.Days % 31 != 0)
         months += 1;
      return $"about {months} {(months == 1 ? "month" : "months")} ago";
   }
   if (span.Days > 0)
      return $"about {span.Days} {(span.Days == 1 ? "day" : "days")} ago";
   if (span.Hours > 0)
      return $"about {span.Hours} {(span.Hours == 1 ? "hour" : "hours")} ago";
   if (span.Minutes > 0)
      return $"about {span.Minutes} {(span.Minutes == 1 ? "minute" : "minutes")} ago";
   if (span.Seconds > 5)
      return $"about {span.Seconds} seconds ago";

   return span.Seconds <= 5 ? "about 5 seconds ago" : string.Empty;
}

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;
    }
public static string RelativeDate(DateTime theDate)
{
    Dictionary<long, string> thresholds = new Dictionary<long, string>();
    int minute = 60;
    int hour = 60 * minute;
    int day = 24 * hour;
    thresholds.Add(60, "{0} seconds ago");
    thresholds.Add(minute * 2, "a minute ago");
    thresholds.Add(45 * minute, "{0} minutes ago");
    thresholds.Add(120 * minute, "an hour ago");
    thresholds.Add(day, "{0} hours ago");
    thresholds.Add(day * 2, "yesterday");
    thresholds.Add(day * 30, "{0} days ago");
    thresholds.Add(day * 365, "{0} months ago");
    thresholds.Add(long.MaxValue, "{0} years ago");
    long since = (DateTime.Now.Ticks - theDate.Ticks) / 10000000;
    foreach (long threshold in thresholds.Keys) 
    {
        if (since < threshold) 
        {
            TimeSpan t = new TimeSpan((DateTime.Now.Ticks - theDate.Ticks));
            return string.Format(thresholds[threshold], (t.Days > 365 ? t.Days / 365 : (t.Days > 0 ? t.Days : (t.Hours > 0 ? t.Hours : (t.Minutes > 0 ? t.Minutes : (t.Seconds > 0 ? t.Seconds : 0))))).ToString());
        }
    }
    return "";
}

我更喜欢这个版本,因为它简洁,并且能够添加新的刻度点。这可以用Timespan的Latest()扩展来封装,而不是长的1行,但为了发布的简洁,这可以。这修复了一小时前、一小时前的问题,提供了一个小时直到两小时过去

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";