计算C#中的相对时间

时间:2008-07-31 23:55:38

标签: c# datetime time datediff relative-time-span

给定特定DateTime值,如何显示相对时间,如:

  • 2小时前
  • 3天前
  • 一个月前

39 个答案:

答案 0 :(得分:932)

Jeff,your code很不错,但可以使用常量更清晰(如Code Complete中所示)。

const int SECOND = 1;
const int MINUTE = 60 * SECOND;
const int HOUR = 60 * MINUTE;
const int DAY = 24 * HOUR;
const int MONTH = 30 * DAY;

var ts = new TimeSpan(DateTime.UtcNow.Ticks - yourDate.Ticks);
double delta = Math.Abs(ts.TotalSeconds);

if (delta < 1 * MINUTE)
  return ts.Seconds == 1 ? "one second ago" : ts.Seconds + " seconds ago";

if (delta < 2 * MINUTE)
  return "a minute ago";

if (delta < 45 * MINUTE)
  return ts.Minutes + " minutes ago";

if (delta < 90 * MINUTE)
  return "an hour ago";

if (delta < 24 * HOUR)
  return ts.Hours + " hours ago";

if (delta < 48 * HOUR)
  return "yesterday";

if (delta < 30 * DAY)
  return ts.Days + " days ago";

if (delta < 12 * 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";
}

答案 1 :(得分:358)

jquery.timeago plugin

Jeff,因为Stack Overflow广泛使用jQuery,我推荐jquery.timeago plugin

优点:

  • 即使页面在10分钟前被打开,也要避免时间戳“1分钟前”; timeago自动刷新。
  • 您可以在Web应用程序中充分利用页面和/或片段缓存,因为时间戳不是在服务器上计算的。
  • 你可以像酷孩子一样使用微格式。

只需将它附加到DOM准备好的时间戳上:

jQuery(document).ready(function() {
    jQuery('abbr.timeago').timeago();
});

这会将所有abbr元素转换为标题类中的timeago和ISO 8601时间戳:

<abbr class="timeago" title="2008-07-17T09:24:17Z">July 17, 2008</abbr>

这样的事情:

<abbr class="timeago" title="July 17, 2008">4 months ago</abbr>

产量:4个月前。随着时间的推移,时间戳将自动更新。

免责声明:我写了这个插件,所以我有偏见。

答案 2 :(得分:326)

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 < 120)
{
  return "a minute ago";
}
if (delta < 2700) // 45 * 60
{
  return ts.Minutes + " minutes ago";
}
if (delta < 5400) // 90 * 60
{
  return "an hour ago";
}
if (delta < 86400) // 24 * 60 * 60
{
  return ts.Hours + " hours ago";
}
if (delta < 172800) // 48 * 60 * 60
{
  return "yesterday";
}
if (delta < 2592000) // 30 * 24 * 60 * 60
{
  return ts.Days + " days ago";
}
if (delta < 31104000) // 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";

连连呢?评论?如何改进这种算法?

答案 3 :(得分:91)

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 "";
}

我更喜欢这个版本的简洁性,以及添加新刻度点的能力。 这可以使用Latest()扩展名来封装Timespan,而不是那个长1衬里,但为了简洁发布,这样做。 这修复了一小时前,1小时前,通过提供一小时直到2小时已经过去

答案 4 :(得分:71)

这里是来自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";
    }
}    

答案 5 :(得分:64)

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;
}

答案 6 :(得分:50)

这是我作为DateTime类的扩展方法添加的实现,它处理未来和过去的日期,并提供一个近似选项,允许您指定您要查找的详细程度(“3小时前”vs“ 3小时23分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();
}

答案 7 :(得分:39)

我建议在客户端计算这个。减少服务器的工作量。

以下是我使用的版本(来自Zach Leatherman)

/*
 * Javascript Humane Dates
 * Copyright (c) 2008 Dean Landolt (deanlandolt.com)
 * Re-write by Zach Leatherman (zachleat.com)
 * 
 * Adopted from the John Resig's pretty.js
 * at http://ejohn.org/blog/javascript-pretty-date
 * and henrah's proposed modification 
 * at http://ejohn.org/blog/javascript-pretty-date/#comment-297458
 * 
 * Licensed under the MIT license.
 */

function humane_date(date_str){
        var time_formats = [
                [60, 'just now'],
                [90, '1 minute'], // 60*1.5
                [3600, 'minutes', 60], // 60*60, 60
                [5400, '1 hour'], // 60*60*1.5
                [86400, 'hours', 3600], // 60*60*24, 60*60
                [129600, '1 day'], // 60*60*24*1.5
                [604800, 'days', 86400], // 60*60*24*7, 60*60*24
                [907200, '1 week'], // 60*60*24*7*1.5
                [2628000, 'weeks', 604800], // 60*60*24*(365/12), 60*60*24*7
                [3942000, '1 month'], // 60*60*24*(365/12)*1.5
                [31536000, 'months', 2628000], // 60*60*24*365, 60*60*24*(365/12)
                [47304000, '1 year'], // 60*60*24*365*1.5
                [3153600000, 'years', 31536000], // 60*60*24*365*100, 60*60*24*365
                [4730400000, '1 century'] // 60*60*24*365*100*1.5
        ];

        var time = ('' + date_str).replace(/-/g,"/").replace(/[TZ]/g," "),
                dt = new Date,
                seconds = ((dt - new Date(time) + (dt.getTimezoneOffset() * 60000)) / 1000),
                token = ' ago',
                i = 0,
                format;

        if (seconds < 0) {
                seconds = Math.abs(seconds);
                token = '';
        }

        while (format = time_formats[i++]) {
                if (seconds < format[0]) {
                        if (format.length == 2) {
                                return format[1] + (i > 1 ? token : ''); // Conditional so we don't return Just Now Ago
                        } else {
                                return Math.round(seconds / format[2]) + ' ' + format[1] + (i > 1 ? token : '');
                        }
                }
        }

        // overflow for centuries
        if(seconds > 4730400000)
                return Math.round(seconds / 4730400000) + ' centuries' + token;

        return date_str;
};

if(typeof jQuery != 'undefined') {
        jQuery.fn.humane_dates = function(){
                return this.each(function(){
                        var date = humane_date(this.title);
                        if(date && jQuery(this).text() != date) // don't modify the dom if we don't have to
                                jQuery(this).text(date);
                });
        };
}

答案 8 :(得分:34)

在Nuget上还有一个名为Humanizer的软件包,它实际上运行得很好

DateTime.UtcNow.AddHours(-30).Humanize() => "yesterday"
DateTime.UtcNow.AddHours(-2).Humanize() => "2 hours ago"

DateTime.UtcNow.AddHours(30).Humanize() => "tomorrow"
DateTime.UtcNow.AddHours(2).Humanize() => "2 hours from now"

TimeSpan.FromMilliseconds(1299630020).Humanize() => "2 weeks"
TimeSpan.FromMilliseconds(1299630020).Humanize(3) => "2 weeks, 1 day, 1 hour"

Scott Hanselman在blog

上写了一篇文章

答案 9 :(得分:29)

@jeff

嘘我的你好像有点长。然而,对“昨天”和“年”的支持似乎更加强大。但根据我的使用经验,这个人最有可能在前30天内查看内容。只有真正的硬核人才会在此之后出现。所以这就是我通常选择保持简短的原因。

这是我目前在我的某个网站上使用的方法。这仅返回相对的日,小时,时间。然后用户必须在输出中“打开”。

public static string ToLongString(this TimeSpan time)
{
    string output = String.Empty;

    if (time.Days > 0)
        output += time.Days + " days ";

    if ((time.Days == 0 || time.Days == 1) && time.Hours > 0)
        output += time.Hours + " hr ";

    if (time.Days == 0 && time.Minutes > 0)
        output += time.Minutes + " min ";

    if (output.Length == 0)
        output += time.Seconds + " sec";

    return output.Trim();
}

答案 10 :(得分:24)

晚会结束了几年,但我要求在过去和将来的日期都这样做,所以我将JeffVincent's合并到此。这是一次三国集团的盛会! :)

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";
            }
        }
    }

答案 11 :(得分:22)

在Java中有一种简单的方法吗? java.util.Date课程似乎相当有限。

这是我快速而又脏的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;
}

答案 12 :(得分:20)

iPhone Objective-C版

+ (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];
    }
}

答案 13 :(得分:19)

鉴于世界和她的丈夫似乎在发布代码示例,这是我刚才写的,基于这些答案。

我特别需要这个代码可以本地化。所以我有两个类 - Grammar,它指定了可本地化的术语,FuzzyDateExtensions,它包含一堆扩展方法。我没有必要处理将来的日期时间,因此没有尝试使用此代码处理它们。

我已经在源代码中留下了一些XMLdoc,但为了简洁起见,删除了大多数(在很明显的地方)。我还没有把每个班级成员都包括在内:

public class Grammar
{
    /// <summary> Gets or sets the term for "just now". </summary>
    public string JustNow { get; set; }
    /// <summary> Gets or sets the term for "X minutes ago". </summary>
    /// <remarks>
    ///     This is a <see cref="String.Format"/> pattern, where <c>{0}</c>
    ///     is the number of minutes.
    /// </remarks>
    public string MinutesAgo { get; set; }
    public string OneHourAgo { get; set; }
    public string HoursAgo { get; set; }
    public string Yesterday { get; set; }
    public string DaysAgo { get; set; }
    public string LastMonth { get; set; }
    public string MonthsAgo { get; set; }
    public string LastYear { get; set; }
    public string YearsAgo { get; set; }
    /// <summary> Gets or sets the term for "ages ago". </summary>
    public string AgesAgo { get; set; }

    /// <summary>
    ///     Gets or sets the threshold beyond which the fuzzy date should be
    ///     considered "ages ago".
    /// </summary>
    public TimeSpan AgesAgoThreshold { get; set; }

    /// <summary>
    ///     Initialises a new <see cref="Grammar"/> instance with the
    ///     specified properties.
    /// </summary>
    private void Initialise(string justNow, string minutesAgo,
        string oneHourAgo, string hoursAgo, string yesterday, string daysAgo,
        string lastMonth, string monthsAgo, string lastYear, string yearsAgo,
        string agesAgo, TimeSpan agesAgoThreshold)
    { ... }
}

FuzzyDateString类包含:

public static class FuzzyDateExtensions
{
    public static string ToFuzzyDateString(this TimeSpan timespan)
    {
        return timespan.ToFuzzyDateString(new Grammar());
    }

    public static string ToFuzzyDateString(this TimeSpan timespan,
        Grammar grammar)
    {
        return GetFuzzyDateString(timespan, grammar);
    }

    public static string ToFuzzyDateString(this DateTime datetime)
    {
        return (DateTime.Now - datetime).ToFuzzyDateString();
    }

    public static string ToFuzzyDateString(this DateTime datetime,
       Grammar grammar)
    {
        return (DateTime.Now - datetime).ToFuzzyDateString(grammar);
    }


    private static string GetFuzzyDateString(TimeSpan timespan,
       Grammar grammar)
    {
        timespan = timespan.Duration();

        if (timespan >= grammar.AgesAgoThreshold)
        {
            return grammar.AgesAgo;
        }

        if (timespan < new TimeSpan(0, 2, 0))    // 2 minutes
        {
            return grammar.JustNow;
        }

        if (timespan < new TimeSpan(1, 0, 0))    // 1 hour
        {
            return String.Format(grammar.MinutesAgo, timespan.Minutes);
        }

        if (timespan < new TimeSpan(1, 55, 0))    // 1 hour 55 minutes
        {
            return grammar.OneHourAgo;
        }

        if (timespan < new TimeSpan(12, 0, 0)    // 12 hours
            && (DateTime.Now - timespan).IsToday())
        {
            return String.Format(grammar.HoursAgo, timespan.RoundedHours());
        }

        if ((DateTime.Now.AddDays(1) - timespan).IsToday())
        {
            return grammar.Yesterday;
        }

        if (timespan < new TimeSpan(32, 0, 0, 0)    // 32 days
            && (DateTime.Now - timespan).IsThisMonth())
        {
            return String.Format(grammar.DaysAgo, timespan.RoundedDays());
        }

        if ((DateTime.Now.AddMonths(1) - timespan).IsThisMonth())
        {
            return grammar.LastMonth;
        }

        if (timespan < new TimeSpan(365, 0, 0, 0, 0)    // 365 days
            && (DateTime.Now - timespan).IsThisYear())
        {
            return String.Format(grammar.MonthsAgo, timespan.RoundedMonths());
        }

        if ((DateTime.Now - timespan).AddYears(1).IsThisYear())
        {
            return grammar.LastYear;
        }

        return String.Format(grammar.YearsAgo, timespan.RoundedYears());
    }
}

我想要实现的关键事项之一,也就是本地化,“今天”只表示“此日历日”,因此IsTodayIsThisMonth,{{1}方法看起来像这样:

IsThisYear

并且舍入方法是这样的(我已经包含public static bool IsToday(this DateTime date) { return date.DayOfYear == DateTime.Now.DayOfYear && date.IsThisYear(); } ,因为它有点不同):

RoundedMonths

我希望人们觉得这很有用和/或有趣:o)

答案 14 :(得分:18)

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();

答案 15 :(得分:17)

在PHP中,我这样做:

<?php
function timesince($original) {
    // array of time period chunks
    $chunks = array(
        array(60 * 60 * 24 * 365 , 'year'),
        array(60 * 60 * 24 * 30 , 'month'),
        array(60 * 60 * 24 * 7, 'week'),
        array(60 * 60 * 24 , 'day'),
        array(60 * 60 , 'hour'),
        array(60 , 'minute'),
    );

    $today = time(); /* Current unix time  */
    $since = $today - $original;

    if($since > 604800) {
    $print = date("M jS", $original);

    if($since > 31536000) {
        $print .= ", " . date("Y", $original);
    }

    return $print;
}

// $j saves performing the count function each time around the loop
for ($i = 0, $j = count($chunks); $i < $j; $i++) {

    $seconds = $chunks[$i][0];
    $name = $chunks[$i][1];

    // finding the biggest chunk (if the chunk fits, break)
    if (($count = floor($since / $seconds)) != 0) {
        break;
    }
}

$print = ($count == 1) ? '1 '.$name : "$count {$name}s";

return $print . " ago";

} ?>

答案 16 :(得分:14)

我以为我会用类和多态来拍摄这个照片。我有一个先前的迭代,它使用了子类,最终有太多的开销。我已经切换到一个更灵活的委托/公共属性对象模型,这是更好的。我的代码非常准确,我希望我能想出一个更好的方法来生成“几个月前”,这似乎并没有过于设计。

我认为我仍然坚持使用Jeff的if-then级联,因为它的代码更少而且更简单(确保它能够按预期工作更容易)。

对于以下代码 PrintRelativeTime.GetRelativeTimeMessage(TimeSpan before)返回相对时间消息(例如“昨天”)。

public class RelativeTimeRange : IComparable
{
    public TimeSpan UpperBound { get; set; }

    public delegate string RelativeTimeTextDelegate(TimeSpan timeDelta);

    public RelativeTimeTextDelegate MessageCreator { get; set; }

    public int CompareTo(object obj)
    {
        if (!(obj is RelativeTimeRange))
        {
            return 1;
        }
        // note that this sorts in reverse order to the way you'd expect, 
        // this saves having to reverse a list later
        return (obj as RelativeTimeRange).UpperBound.CompareTo(UpperBound);
    }
}

public class PrintRelativeTime
{
    private static List<RelativeTimeRange> timeRanges;

    static PrintRelativeTime()
    {
        timeRanges = new List<RelativeTimeRange>{
            new RelativeTimeRange
            {
                UpperBound = TimeSpan.FromSeconds(1),
                MessageCreator = (delta) => 
                { return "one second ago"; }
            }, 
            new RelativeTimeRange
            {
                UpperBound = TimeSpan.FromSeconds(60),
                MessageCreator = (delta) => 
                { return delta.Seconds + " seconds ago"; }

            }, 
            new RelativeTimeRange
            {
                UpperBound = TimeSpan.FromMinutes(2),
                MessageCreator = (delta) => 
                { return "one minute ago"; }
            }, 
            new RelativeTimeRange
            {
                UpperBound = TimeSpan.FromMinutes(60),
                MessageCreator = (delta) => 
                { return delta.Minutes + " minutes ago"; }
            }, 
            new RelativeTimeRange
            {
                UpperBound = TimeSpan.FromHours(2),
                MessageCreator = (delta) => 
                { return "one hour ago"; }
            }, 
            new RelativeTimeRange
            {
                UpperBound = TimeSpan.FromHours(24),
                MessageCreator = (delta) => 
                { return delta.Hours + " hours ago"; }
            }, 
            new RelativeTimeRange
            {
                UpperBound = TimeSpan.FromDays(2),
                MessageCreator = (delta) => 
                { return "yesterday"; }
            }, 
            new RelativeTimeRange
            {
                UpperBound = DateTime.Now.Subtract(DateTime.Now.AddMonths(-1)),
                MessageCreator = (delta) => 
                { return delta.Days + " days ago"; }
            }, 
            new RelativeTimeRange
            {
                UpperBound = DateTime.Now.Subtract(DateTime.Now.AddMonths(-2)),
                MessageCreator = (delta) => 
                { return "one month ago"; }
            }, 
            new RelativeTimeRange
            {
                UpperBound = DateTime.Now.Subtract(DateTime.Now.AddYears(-1)),
                MessageCreator = (delta) => 
                { return (int)Math.Floor(delta.TotalDays / 30) + " months ago"; }
            }, 
            new RelativeTimeRange
            {
                UpperBound = DateTime.Now.Subtract(DateTime.Now.AddYears(-2)),
                MessageCreator = (delta) => 
                { return "one year ago"; }
            }, 
            new RelativeTimeRange
            {
                UpperBound = TimeSpan.MaxValue,
                MessageCreator = (delta) => 
                { return (int)Math.Floor(delta.TotalDays / 365.24D) + " years ago"; }
            }
        };

        timeRanges.Sort();
    }

    public static string GetRelativeTimeMessage(TimeSpan ago)
    {
        RelativeTimeRange postRelativeDateRange = timeRanges[0];

        foreach (var timeRange in timeRanges)
        {
            if (ago.CompareTo(timeRange.UpperBound) <= 0)
            {
                postRelativeDateRange = timeRange;
            }
        }

        return postRelativeDateRange.MessageCreator(ago);
    }
}

答案 17 :(得分:12)

你可以试试这个。我认为它会正常工作。

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";
}

答案 18 :(得分:12)

当您知道观众的时区时,可能会更清楚地使用日期等级的日历日。我不熟悉.NET库,所以我不知道你在C#中是怎么做的。不幸的是。

在消费者网站上,你也可能在一分钟之内变得更加手淫。 “不到一分钟前”或“刚才”可能已经足够好了。

答案 19 :(得分:12)

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;
    }
}

another answer to this question相同,但作为带静态字典的扩展方法。

答案 20 :(得分:10)

@Jeff

var ts = new TimeSpan(DateTime.UtcNow.Ticks - dt.Ticks);

DateTime上进行减法无论如何都会返回TimeSpan

所以你可以做到

(DateTime.UtcNow - dt).TotalSeconds

我也很惊讶地发现常数是手工增加的,然后在乘法中添加了注释。这是一些错误的优化吗?

答案 21 :(得分:9)

用于客户端gwt使用的Java:

import java.util.Date;

public class RelativeDateFormat {

 private static final long ONE_MINUTE = 60000L;
 private static final long ONE_HOUR = 3600000L;
 private static final long ONE_DAY = 86400000L;
 private static final long ONE_WEEK = 604800000L;

 public static String format(Date date) {

  long delta = new Date().getTime() - date.getTime();
  if (delta < 1L * ONE_MINUTE) {
   return toSeconds(delta) == 1 ? "one second ago" : toSeconds(delta)
     + " seconds ago";
  }
  if (delta < 2L * ONE_MINUTE) {
   return "one minute ago";
  }
  if (delta < 45L * ONE_MINUTE) {
   return toMinutes(delta) + " minutes ago";
  }
  if (delta < 90L * ONE_MINUTE) {
   return "one hour ago";
  }
  if (delta < 24L * ONE_HOUR) {
   return toHours(delta) + " hours ago";
  }
  if (delta < 48L * ONE_HOUR) {
   return "yesterday";
  }
  if (delta < 30L * ONE_DAY) {
   return toDays(delta) + " days ago";
  }
  if (delta < 12L * 4L * ONE_WEEK) {
   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 static long toSeconds(long date) {
  return date / 1000L;
 }

 private static long toMinutes(long date) {
  return toSeconds(date) / 60L;
 }

 private static long toHours(long date) {
  return toMinutes(date) / 60L;
 }

 private static long toDays(long date) {
  return toHours(date) / 24L;
 }

 private static long toMonths(long date) {
  return toDays(date) / 30L;
 }

 private static long toYears(long date) {
  return toMonths(date) / 365L;
 }

}

答案 22 :(得分:8)

这是算法stackoverflow使用但在perlish伪代码中更简洁地重写并修复了错误(没有“一小时前”)。该函数需要一个(正)秒数,并返回一个人性化的字符串,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";

答案 23 :(得分:8)

您可以使用TimeAgo extension,其中如下所示:

public static string TimeAgo(this DateTime dateTime)
{
    string result = string.Empty;
    var timeSpan = DateTime.Now.Subtract(dateTime);

    if (timeSpan <= TimeSpan.FromSeconds(60))
    {
        result = string.Format("{0} seconds ago", timeSpan.Seconds);
    }
    else if (timeSpan <= TimeSpan.FromMinutes(60))
    {
        result = timeSpan.Minutes > 1 ? 
            String.Format("about {0} minutes ago", timeSpan.Minutes) :
            "about a minute ago";
    }
    else if (timeSpan <= TimeSpan.FromHours(24))
    {
        result = timeSpan.Hours > 1 ? 
            String.Format("about {0} hours ago", timeSpan.Hours) : 
            "about an hour ago";
    }
    else if (timeSpan <= TimeSpan.FromDays(30))
    {
        result = timeSpan.Days > 1 ? 
            String.Format("about {0} days ago", timeSpan.Days) : 
            "yesterday";
    }
    else if (timeSpan <= TimeSpan.FromDays(365))
    {
        result = timeSpan.Days > 30 ? 
            String.Format("about {0} months ago", timeSpan.Days / 30) : 
            "about a month ago";
    }
    else
    {
        result = timeSpan.Days > 365 ? 
            String.Format("about {0} years ago", timeSpan.Days / 365) : 
            "about a year ago";
    }

    return result;
}

或者使用来自Timeago的Razor扩展程序jQuery plugin

答案 24 :(得分:8)

这是我从比尔盖茨的一个博客那里得到的。我需要在浏览器历史记录中找到它,然后我会给你链接。

Javascript代码执行相同的操作(根据要求):

function posted(t) {
    var now = new Date();
    var diff = parseInt((now.getTime() - Date.parse(t)) / 1000);
    if (diff < 60) { return 'less than a minute ago'; }
    else if (diff < 120) { return 'about a minute ago'; }
    else if (diff < (2700)) { return (parseInt(diff / 60)).toString() + ' minutes ago'; }
    else if (diff < (5400)) { return 'about an hour ago'; }
    else if (diff < (86400)) { return 'about ' + (parseInt(diff / 3600)).toString() + ' hours ago'; }
    else if (diff < (172800)) { return '1 day ago'; } 
    else {return (parseInt(diff / 86400)).toString() + ' days ago'; }
}

基本上,你的工作时间是几秒钟......

答案 25 :(得分:8)

您可以通过执行此逻辑客户端来减少服务器端负载。查看一些Digg页面上的源代码以供参考。他们让服务器发出一个由Javascript处理的纪元时间值。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());
}

你甚至可以在那里添加NOSCRIPT块,只需执行ToString()。

答案 26 :(得分:6)

/** 
 * {@code date1} has to be earlier than {@code date2}.
 */
public static String relativize(Date date1, Date date2) {
    assert date2.getTime() >= date1.getTime();

    long duration = date2.getTime() - date1.getTime();
    long converted;

    if ((converted = TimeUnit.MILLISECONDS.toDays(duration)) > 0) {
        return String.format("%d %s ago", converted, converted == 1 ? "day" : "days");
    } else if ((converted = TimeUnit.MILLISECONDS.toHours(duration)) > 0) {
        return String.format("%d %s ago", converted, converted == 1 ? "hour" : "hours");
    } else if ((converted = TimeUnit.MILLISECONDS.toMinutes(duration)) > 0) {
        return String.format("%d %s ago", converted, converted == 1 ? "minute" : "minutes");
    } else if ((converted = TimeUnit.MILLISECONDS.toSeconds(duration)) > 0) {
        return String.format("%d %s ago", converted, converted == 1 ? "second" : "seconds");
    } else {
        return "just now";
    }
}

答案 27 :(得分:6)

如果您希望获得TimeSpan timeDiff = DateTime.Now-CreatedDate; 之类的输出,则需要一个时间跨度:

timeDiff.Days
timeDiff.Hours

然后你可以访问你喜欢的值:

{{1}}

等...

答案 28 :(得分:6)

我认为已经有很多与这篇文章相关的答案,但是可以使用它,它就像插件一样易于使用,也易于程序员阅读。 发送您的具体日期,并以字符串形式获取其值:

public string RelativeDateTimeCount(DateTime inputDateTime)
{
    string outputDateTime = string.Empty;
    TimeSpan ts = DateTime.Now - inputDateTime;

    if (ts.Days > 7)
    { outputDateTime = inputDateTime.ToString("MMMM d, yyyy"); }

    else if (ts.Days > 0)
    {
        outputDateTime = ts.Days == 1 ? ("about 1 Day ago") : ("about " + ts.Days.ToString() + " Days ago");
    }
    else if (ts.Hours > 0)
    {
        outputDateTime = ts.Hours == 1 ? ("an hour ago") : (ts.Hours.ToString() + " hours ago");
    }
    else if (ts.Minutes > 0)
    {
        outputDateTime = ts.Minutes == 1 ? ("1 minute ago") : (ts.Minutes.ToString() + " minutes ago");
    }
    else outputDateTime = "few seconds ago";

    return outputDateTime;
}

答案 29 :(得分:5)

var ts = new TimeSpan(DateTime.Now.Ticks - dt.Ticks);

答案 30 :(得分:4)

public string getRelativeDateTime(DateTime date)
{
    TimeSpan ts = DateTime.Now - date;
    if (ts.TotalMinutes < 1)//seconds ago
        return "just now";
    if (ts.TotalHours < 1)//min ago
        return (int)ts.TotalMinutes == 1 ? "1 Minute ago" : (int)ts.TotalMinutes + " Minutes ago";
    if (ts.TotalDays < 1)//hours ago
        return (int)ts.TotalHours == 1 ? "1 Hour ago" : (int)ts.TotalHours + " Hours ago";
    if (ts.TotalDays < 7)//days ago
        return (int)ts.TotalDays == 1 ? "1 Day ago" : (int)ts.TotalDays + " Days ago";
    if (ts.TotalDays < 30.4368)//weeks ago
        return (int)(ts.TotalDays / 7) == 1 ? "1 Week ago" : (int)(ts.TotalDays / 7) + " Weeks ago";
    if (ts.TotalDays < 365.242)//months ago
        return (int)(ts.TotalDays / 30.4368) == 1 ? "1 Month ago" : (int)(ts.TotalDays / 30.4368) + " Months ago";
    //years ago
    return (int)(ts.TotalDays / 365.242) == 1 ? "1 Year ago" : (int)(ts.TotalDays / 365.242) + " Years ago";
}

一个月和一年中的天数转换值来自Google。

答案 31 :(得分:4)

我会为此提供一些方便的扩展方法,并使代码更具可读性。首先,Int32的几种扩展方法。

public static class TimeSpanExtensions {

    public static TimeSpan Days(this int value) {

        return new TimeSpan(value, 0, 0, 0);
    }

    public static TimeSpan Hours(this int value) {

        return new TimeSpan(0, value, 0, 0);
    }

    public static TimeSpan Minutes(this int value) {

        return new TimeSpan(0, 0, value, 0);
    }

    public static TimeSpan Seconds(this int value) {

        return new TimeSpan(0, 0, 0, value);
    }

    public static TimeSpan Milliseconds(this int value) {

        return new TimeSpan(0, 0, 0, 0, value);
    }

    public static DateTime Ago(this TimeSpan value) {

        return DateTime.Now - value;
    }
}

然后,一个用于DateTime

public static class DateTimeExtensions {

    public static DateTime Ago(this DateTime dateTime, TimeSpan delta) {

        return dateTime - delta;
    }
}

现在,您可以执行以下操作:

var date = DateTime.Now;
date.Ago(2.Days()); // 2 days ago
date.Ago(7.Hours()); // 7 hours ago
date.Ago(567.Milliseconds()); // 567 milliseconds ago

答案 32 :(得分:2)

当然,解决“1小时前”问题的一个简单方法是增加“一小时前”有效的窗口。 变化

if (delta < 5400) // 90 * 60
{
    return "an hour ago";
}

if (delta < 7200) // 120 * 60
{
    return "an hour ago";
}

这意味着110分钟前发生的事情会在“一小时前”读到 - 这可能并不完美,但我认为这比“1小时前”的情况要好。

答案 33 :(得分:2)

在通过计算相对时间(从几秒钟到几年)来完成DateTime函数的方式中,请尝试如下操作:

using System;

public class Program {
    public static string getRelativeTime(DateTime past) {
        DateTime now = DateTime.Today;
        string rt = "";
        int time;
        string statement = "";
        if (past.Second >= now.Second) {
            if (past.Second - now.Second == 1) {
                rt = "second ago";
            }
            rt = "seconds ago";
            time = past.Second - now.Second;
            statement = "" + time;
            return (statement + rt);
        }
        if (past.Minute >= now.Minute) {
            if (past.Second - now.Second == 1) {
                rt = "second ago";
            } else {
                rt = "minutes ago";
            }
            time = past.Minute - now.Minute;
            statement = "" + time;
            return (statement + rt);
        }
        // This process will go on until years
    }
    public static void Main() {
        DateTime before = new DateTime(1995, 8, 24);
        string date = getRelativeTime(before);
        Console.WriteLine("Windows 95 was {0}.", date);
    }
}

不能完全正常工作,但是如果您对其进行一些修改和调试,则可能会完成这项工作。

答案 34 :(得分:1)

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;
        }

答案 35 :(得分:0)

我的方式更简单。您可以根据需要调整返回字符串

    public static string TimeLeft(DateTime utcDate)
    {
        TimeSpan timeLeft = DateTime.UtcNow - utcDate;
        string timeLeftString = "";
        if (timeLeft.Days > 0)
        {
            timeLeftString += timeLeft.Days == 1 ? timeLeft.Days + " day" : timeLeft.Days + " days";
        }
        else if (timeLeft.Hours > 0)
        {
            timeLeftString += timeLeft.Hours == 1 ? timeLeft.Hours + " hour" : timeLeft.Hours + " hours";
        }
        else
        {
            timeLeftString += timeLeft.Minutes == 1 ? timeLeft.Minutes+" minute" : timeLeft.Minutes + " minutes";
        }
        return timeLeftString;
    }

答案 36 :(得分:0)

土耳其语的Vincents答案的本地化版本。

    const int SECOND = 1;
    const int MINUTE = 60 * SECOND;
    const int HOUR = 60 * MINUTE;
    const int DAY = 24 * HOUR;
    const int MONTH = 30 * DAY;

    var ts = new TimeSpan(DateTime.UtcNow.Ticks - yourDate.Ticks);
    double delta = Math.Abs(ts.TotalSeconds);

    if (delta < 1 * MINUTE)
        return ts.Seconds + " saniye önce";

    if (delta < 45 * MINUTE)
        return ts.Minutes + " dakika önce";

    if (delta < 24 * HOUR)
        return ts.Hours + " saat önce";

    if (delta < 48 * HOUR)
        return "dün";

    if (delta < 30 * DAY)
        return ts.Days + " gün önce";

    if (delta < 12 * MONTH)
    {
        int months = Convert.ToInt32(Math.Floor((double)ts.Days / 30));
        return months + " ay önce";
    }
    else
    {
        int years = Convert.ToInt32(Math.Floor((double)ts.Days / 365));
        return years + " yıl önce";
    }

答案 37 :(得分:0)

// Calculate total days in current year
int daysInYear;

for (var i = 1; i <= 12; i++)
    daysInYear += DateTime.DaysInMonth(DateTime.Now.Year, i);

// Past date
DateTime dateToCompare = DateTime.Now.Subtract(TimeSpan.FromMinutes(582));

// Calculate difference between current date and past date
double diff = (DateTime.Now - dateToCompare).TotalMilliseconds;

TimeSpan ts = TimeSpan.FromMilliseconds(diff);

var years = ts.TotalDays / daysInYear; // Years
var months = ts.TotalDays / (daysInYear / (double)12); // Months
var weeks = ts.TotalDays / 7; // Weeks
var days = ts.TotalDays; // Days
var hours = ts.TotalHours; // Hours
var minutes = ts.TotalMinutes; // Minutes
var seconds = ts.TotalSeconds; // Seconds

if (years >= 1)
    Console.WriteLine(Math.Round(years, 0) + " year(s) ago");
else if (months >= 1)
    Console.WriteLine(Math.Round(months, 0) + " month(s) ago");
else if (weeks >= 1)
    Console.WriteLine(Math.Round(weeks, 0) + " week(s) ago");
else if (days >= 1)
    Console.WriteLine(Math.Round(days, 0) + " days(s) ago");
else if (hours >= 1)
    Console.WriteLine(Math.Round(hours, 0) + " hour(s) ago");
else if (minutes >= 1)
    Console.WriteLine(Math.Round(minutes, 0) + " minute(s) ago");
else if (seconds >= 1)
    Console.WriteLine(Math.Round(seconds, 0) + " second(s) ago");

Console.ReadLine();

答案 38 :(得分:0)

使用解构和Linq获取“ n [最大时间单位]之前”的“单线”:

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");

您得到786 year(s) ago

与当前年份和月份类似,例如

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

您得到4 day(s) ago

带有实际日期,例如

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

您得到9 hour(s) ago