Question

Possible Duplicate:
Sorting objects in an array by a field value in JavaScript

How can I sort an array of objects numerically (by id) then alphabetically (by name)?

The current way is providing invalid output.

This is the object i’m trying to sort through

var items = [
    {
        "id": 165,
        "name": "a"
    },
    {
        "id": 236,
        "name": "c"
    },
    {
        "id": 376,
        "name": "b"
    },
    {
        "id": 253,
        "name": "f"
    },
    {
        "id": 235,
        "name": "e"
    },
    {
        "id": 24,
        "name": "d"
    },
    {
        "id": 26,
        "name": "d"
    }
]

and the way i’m trying to sort

items.sort(function(a, b) {
    return (a.id - b.id);
}).sort(function(a, b) {
    return (a.name - b.name);
});

here is the jsfiddle.

http://jsfiddle.net/jh4xb/

EDIT: Sorry for the confusion, I’ve been so confused by this problem for a while.

What I’m trying to accomplish is to sort by the highest id first, and then sort alphabetically so in the end it should look like:

var items = [
    {
        "id": 376,
        "name": "b"
    },
    {
        "id": 253,
        "name": "f"
    },
    {
        "id": 236,
        "name": "c"
    },
    {
        "id": 235,
        "name": "e"
    },
    {
        "id": 165,
        "name": "a"
    },
    {
        "id": 26,
        "name": "d"
    },
    {
        "id": 24,
        "name": "d"
    }
]

Answer 1

I think it’s better done just with…

items.sort(function(a, b) { 
  return a.id - b.id  ||  a.name.localeCompare(b.name);
});

The second sort will basically negate the first one, so you have to do it once. )

a.id - b.id || a.name.localeCompare(b.name) expression will first compare the ids; only if they are equal, it will compare the names (and return the result of this comparison).

If you need to reverse the ordering, swap the positions (b.id - a.id, etc.) – or just negate the whole thing:

items.sort(function(a, b) { 
  return - ( a.id - b.id  ||  a.name.localeCompare(b.name) );
});

Here’s the JSFiddle (have to simplify the data a bit to show my point).