current community

your communities

more stack exchange communities

company blog
Stack Exchange Inbox Reputation and Badges
tour help
Sign up ×
Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute:
up vote 2 down vote favorite

I have a collection objects in an array,

[{A:2, B:2},{A:2, B:3},{A:1, B: 2}, {A:1, B:1}]

I would like to have a sort result of property A then property B. Therefore, as a result it should be: 

[{A:1, B:1}, {A:1, B:2}, {A:2, B:2}, {A:2, B:3}]
share|improve this question
    
possible duplicate of How to sort a JSON array? – luvieere Sep 8 '11 at 9:47
up vote 4 down vote accepted

Sort by A then by B. This sort works, because if A == A then A - A is 0. And 0 || anything is always anything. So it compares A then differs to B is A's are equal.

var list = [{A:2, B:2},{A:2, B:3},{A:1, B: 2}, {A:1, B:1}];
list.sort(function (a, b) {
  return a.A - b.A || a.B - b.B
})

Using fully cross browser sort and the benefits of lazy evaluation

share|improve this answer
    
Great! thanks you so much for your time. – Vicheanak Sep 7 '11 at 4:34
    
Nice edit of your original post which didn't accomplish the desired result (which is why I wrote my answer). – jfriend00 Sep 7 '11 at 4:40
up vote 1 down vote

try following code

[{A:2, B:2},{A:2, B:3},{A:1, B: 2}, {A:1, B:1}]
.sort(function(a, b) { return (a.A< b.A) ? -1 : (a.A> b.A) ? 1 : 0 })
.sort(function(a, b) { return (a.B< b.B) ? -1 : (a.B> b.B) ? 1 : 0 });
share|improve this answer
up vote 3 down vote

You use a custom sort callback that first compares the A values and then, if those are equal uses the B values, thus creating primary and secondary keys for the sort:

var collection = [{A:2, B:2},{A:2, B:3},{A:1, B: 2}, {A:1, B:1}];

collection.sort(function(a, b){
    if (a.A != b.A) {
        return(a.A - b.a);
    } else {
        return(a.B - b.B);
    }
});
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.