如何在 JavaScript 中呈现两个时间戳之间的上下文差异

Question

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'

或

'in about 15 minutes'

（应该支持过去和未来）。

过去的差异有解决方案，但我还没有找到支持未来时间差异的解决方案。

这些是我尝试过的解决方案：

约翰·雷西格的《漂亮约会》和扎克·莱瑟曼的修改

jQuery解决方案的奖励积分。

Answer 1

查看您链接的解决方案...这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，它为你将来的日期加上"In"。如您所见，更改非常小。

  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

呵

呵 - 我昨天实际上写了一个函数来做这件事（它不在这台计算机上，所以我只需要试着记住它）

我扩展了 Date 原型类，但这可以很容易地放入常规函数中。

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