如何在JavaScript中呈现两个时间戳之间的上下文差异？

假设我在JavaScript中有两个字符串：

var date1 = '2008-10-03T20:24Z' var date2 = '2008-10-04T12:24Z' 

我怎么会得到这样的结果：

 '4 weeks ago' 

要么

 'in about 15 minutes' 

（应该支持过去和未来）。

过去的差异有解决方案，但我还没有找到一个支持未来时间差异的解决方案。

这些是我尝试的解决方案：

John Resig的漂亮日期和Zach Leatherman的修改

jQuery解决方案的奖励积分。

Answer 1

看看你链接的解决方案……它实际上就像我轻浮的评论一样简单！

这是Zach Leatherman代码的一个版本，它代表“In”为您预定未来的日期。 如您所见，这些变化非常小。

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

Answer 2

嘿 – 我昨天写了一个函数来做这件事（这不是在这台电脑上，所以我只需要试着记住它）

我扩展了Date原型类，但这很容易被放入常规函数中。

 Date.prototype.toRelativeTime = function(otherTime) { // if no parameter is passed, use the current date. if (otherTime == undefined) otherTime = new Date(); var diff = Math.abs(this.getTime() - otherTime.getTime()) / 1000; var MIN = 60, // some "constants" just HOUR = 3600, // for legibility DAY = 86400 ; var out, temp; if (diff < MIN) { out = "Less than a minute"; } else if (diff < 15 * MIN) { // less than fifteen minutes, show how many minutes temp = Math.round(diff / MIN); out = temp + " minute" + (temp == 1 ? "" : "s"); // eg: 12 minutes } else if (diff < HOUR) { // less than an hour, round down to the nearest 5 minutes out = (Math.floor(diff / (5 * MIN)) * 5) + " minutes"; } else if (diff < DAY) { // less than a day, just show hours temp = Math.round(diff / HOUR); out = temp + " hour" + (temp == 1 ? "" : "s"); } else if (diff < 30 * DAY) { // show how many days ago temp = Math.round(diff / DAY); out = temp + " day" + (temp == 1 ? "" : "s"); } else if (diff < 90 * DAY) { // more than 30 days, but less than 3 months, show the day and month return this.getDate() + " " + this.getShortMonth(); // see below } else { // more than three months difference, better show the year too return this.getDate() + " " + this.getShortMonth() + " " + this.getFullYear(); } return out + (this.getTime() > otherTime.getTime() ? " from now" : " ago"); }; Date.prototype.getShortMonth = function() { return ["Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"][this.getMonth()]; }; // sample usage: var x = new Date(2008, 9, 4, 17, 0, 0); alert(x.toRelativeTime()); // 9 minutes from now x = new Date(2008, 9, 4, 16, 45, 0, 0); alert(x.toRelativeTime()); // 6 minutes ago x = new Date(2008, 11, 1); // 1 Dec x = new Date(2009, 11, 1); // 1 Dec 2009