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

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];
    }
}

其他回答

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";

Nuget上还有一个名为Humanizr的软件包,它实际上运行得很好,并且在.NET Foundation中。

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在他的博客上写了一篇文章

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);
                });
        };
}

@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?

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];
    }
}