Sort array by two values [duplicate]

Question

Possible Duplicate:
Simple function to sort an array of objects

While writing the script, there was a problem with sorting an array. I have an array of objects:

[{gold: false, range: 1601}, {gold: true, range: 13}, {gold: false, range: 375}, {gold: true, range: 601}

But it should look like this:

[{gold: true, range: 13}, {gold: true, range: 601}, {gold:false, range: 375}, {gold: false, range: 1601}]

I want to get an array in which the keys are sorted by increasing range. But if there gold key value is true, then they are the first.

explain your problem in details

Manish Nagar
– Manish Nagar

2012-12-29 12:27:21 +00:00
Commented Dec 29, 2012 at 12:27 — Manish Nagar
– Manish Nagar, Commented Dec 29, 2012 at 12:27

SergeyS · Accepted Answer · 2012-12-29 12:50:44Z

11

Use something like this:

yourArray.sort(function(a, b){
    return a["gold"] == b["gold"] ? a["range"] - b["range"] : a["gold"] ? -1 : 1;
});

Or even this:

yourArray.sort(function(a, b){
    return  b["gold"] - a["gold"] || a["range"] - b["range"];
});

The second approach is really cool actually)) You can just use this pattern for any number of fields in your object, just order them by their importance. If some field should be sorted as ascending - than a["..."] - b["...], if as descending - than b["..."] - a["..."]

edited Dec 29, 2012 at 12:50

answered Dec 29, 2012 at 12:32

SergeyS

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Yury Tarabanko · Accepted Answer · 2012-12-29 12:28:38Z

3

Try

var arr = [{gold: false, range: 1601}, {gold: true, range: 13}, {gold: false, range: 375}, {gold: true, range: 601}], 
    sorter = function(a, b) {
      if(a.gold && !b.gold) {return -1;}
      if(!a.gold && b.gold) {return 1;}
      return a.range - b.range;
    };

arr.sort(sorter);

answered Dec 29, 2012 at 12:28

Yury Tarabanko

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