Sort array by ISO 8601 date

Question

how can i sort this array by date (ISO 8601)?

var myArray = new Array();  myArray[0] = { name:'oldest', date:'2007-01-17T08:00:00Z' } myArray[1] = { name:'newest', date:'2011-01-28T08:00:00Z' } myArray[2] = { name:'old',    date:'2009-11-25T08:00:00Z' } 

Playground:
http://jsfiddle.net/4tUZt/

Thanks in advance!

Answer 1

Sort Lexicographically:

As @kdbanman points out, ISO8601^{See General principles} was designed for lexicographical sort. As such the ISO8601 string representation can be sorted like any other string, and this will give the expected order.

'2007-01-17T08:00:00Z' < '2008-01-17T08:00:00Z' === true 

So you would implement:

var myArray = [     { name:'oldest', date:'2007-01-17T08:00:00Z' },     { name:'newest', date:'2011-01-28T08:00:00Z' },     { name:'old',    date:'2009-11-25T08:00:00Z' } ];  myArray.sort(function(a, b) {     return (a.date < b.date) ? -1 : ((a.date > b.date) ? 1 : 0); }); 

---

Sort using JavaScript Date:

Older versions of WebKit and Internet Explorer do not support ISO 8601 dates, so you have to make a compatible date. It is supported by FireFox, and modern WebKit though See here for more information about Date.parse support JavaScript: Which browsers support parsing of ISO-8601 Date String with Date.parse

Here is a very good article for creating a Javascript ISO 8601 compatible date, which you can then sort like regular javascript dates.

http://webcloud.se/log/JavaScript-and-ISO-8601/

Date.prototype.setISO8601 = function (string) {     var regexp = "([0-9]{4})(-([0-9]{2})(-([0-9]{2})" +     "(T([0-9]{2}):([0-9]{2})(:([0-9]{2})(\.([0-9]+))?)?" +     "(Z|(([-+])([0-9]{2}):([0-9]{2})))?)?)?)?";     var d = string.match(new RegExp(regexp));      var offset = 0;     var date = new Date(d[1], 0, 1);      if (d[3]) { date.setMonth(d[3] - 1); }     if (d[5]) { date.setDate(d[5]); }     if (d[7]) { date.setHours(d[7]); }     if (d[8]) { date.setMinutes(d[8]); }     if (d[10]) { date.setSeconds(d[10]); }     if (d[12]) { date.setMilliseconds(Number("0." + d[12]) * 1000); }     if (d[14]) {         offset = (Number(d[16]) * 60) + Number(d[17]);         offset *= ((d[15] == '-') ? 1 : -1);     }      offset -= date.getTimezoneOffset();     time = (Number(date) + (offset * 60 * 1000));     this.setTime(Number(time)); } 

Usage:

console.log(myArray.sort(sortByDate));    function sortByDate( obj1, obj2 ) {     var date1 = (new Date()).setISO8601(obj1.date);     var date2 = (new Date()).setISO8601(obj2.date);     return date2 > date1 ? 1 : -1; } 

Updated usage to include sorting technique credit @nbrooks

Answer 2

You can avoid creating of dates and by using the built–in lexicographic compare function String.prototype.localeCompare, rather than the ?: compound operator or other expressions:

var myArray = [    {name: 'oldest', date: '2007-01-17T08:00:00Z'},    {name: 'newest', date: '2011-01-28T08:00:00Z'},    {name: 'old', date: '2009-11-25T08:00:00Z'}  ];    // Oldest first  console.log(    myArray.sort((a, b) => a.date.localeCompare(b.date))  );    // Newest first  console.log(    myArray.sort((a, b) => -a.date.localeCompare(b.date))  );