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

鉴于全世界和她的丈夫似乎都在发布代码样本,这是我不久前根据这些答案写的。

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,它包含一系列扩展方法。我不需要处理未来的日期时间,因此不尝试使用此代码处理它们。

为了简洁起见,我在源代码中保留了一些XMLdoc,但删除了大部分(显而易见的地方)。我也没有把每个班级成员都包括在这里:

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());
    }
}

我想实现的一件关键事情,以及本地化,就是“今天”只意味着“这个日历日”,所以IsToday、IsThisMonth和IsThisYear方法如下:

public static bool IsToday(this DateTime date)
{
    return date.DayOfYear == DateTime.Now.DayOfYear && date.IsThisYear();
}

舍入方法如下(我已包括RoundedMonths,因为这有点不同):

public static int RoundedDays(this TimeSpan timespan)
{
    return (timespan.Hours > 12) ? timespan.Days + 1 : timespan.Days;
}

public static int RoundedMonths(this TimeSpan timespan)
{
    DateTime then = DateTime.Now - timespan;

    // Number of partial months elapsed since 1 Jan, AD 1 (DateTime.MinValue)
    int nowMonthYears = DateTime.Now.Year * 12 + DateTime.Now.Month;
    int thenMonthYears = then.Year * 12 + then.Month;                    

    return nowMonthYears - thenMonthYears;
}

我希望人们觉得这很有用和/或有趣:o)

其他回答

我认为已经有很多关于这篇文章的答案了,但你可以使用它,它就像插件一样容易使用,程序员也很容易阅读。发送您的特定日期,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;
}

我也建议在客户端进行计算。服务器工作更少。

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);
                });
        };
}

如果您想获得类似“2天4小时12分钟前”的输出,则需要一个时间跨度:

TimeSpan timeDiff = DateTime.Now-CreatedDate;

然后您可以访问您喜欢的值:

timeDiff.Days
timeDiff.Hours

@杰夫

var ts=新时间跨度(DateTime.UtcNow.Ticks-dt.Ticks);

对DateTime执行减法仍会返回TimeSpan。

所以你可以这样做

(DateTime.UtcNow - dt).TotalSeconds

我也很惊讶地看到常数用手相乘,然后注释加上乘法。这是错误的优化吗?

在Java中有没有一种简单的方法可以做到这一点?java.util.Date类似乎相当有限。

下面是我的快速而肮脏的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;
}