sorting an object respectively array effectively with jquery

Question

I am interested in how you would sort the following data most effectively with jQuery:

There is a json object which I retrieve from the server:

var data = {/*json-object*/};
$.each(data.level1.sub1, function(){
    if(this.name.length < 2) {
        this.type = "banana";
        itemarray.push(this);
    }
    else {
        this.type = "apple";
        itemarray.push(this);
    }
});

$.each(data.level2.sub1, function(){
    this.type = "kiwi";
    itemarray.push(this);
});

This brings me to the point that I can list what's in the array (which is typeof object, not a "real" array) by doing this:

if (itemarray.length){
    $.each(itemarray, function(index, value){
      $('#wrapper').append('<li>' + index + ' : ' + value.name + ' * ' + value.type + '</li>');
    });
}

Result:

0 : abc * apple
1 : a * banana
2 : lala * apple
3 : bcd - Copy * kiwi
4 : okl * banana

Now I would like to sort the array (which is typeof object, again): group by type and sort asc per type. How would you suggest doing that?

Rob W · Accepted Answer · 2011-11-24 11:03:48Z

1

Since you're using the itemarray.push method, I assume that itemarray is a true array. Array objects (note: typeof [] === "object") support a sort() method.

Use the native Array.sort method, in conjunction with a function, as shown below. This method directly modifies the array.

var itemarray = [{type:"C"},{type: "B"}]; //Your array looks like this

//ASC result: [{type:"C"}, {type:"B"}]
itemarray.sort(function(x,y){
    return x.type.localeCompare(y.type);
});
//DESC result: [{type:"B"}, {type: "B"}]
itemarray.sort(function(x,y){
    return y.type.localeCompare(x.type);
});

answered Nov 24, 2011 at 11:03

Rob W

350k87 gold badges811 silver badges683 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

ho.s Over a year ago

hey rob, thanks a lot so far. your sorting either brings me to:

0 : aname * apple 1 : bname * apple 2 : cname * apple 3 : aname * banana 4 : bname * banana 2 : aname * kiwi 3 : bname * kiwi

when i process the first sorting method twice (x.type.localeCompare(y.type) not the other way round), thats nice, because it sorts types and names each. BUT i actually would like to have the order: apple, kiwi, banana! I tried out a few combinations, but sort() sorts the types-names themselves - so how can i get to this?

ho.s Over a year ago

ok, got it: to group the type-groups in the desired order, i do the following after sorting:

applearray = $.grep(itemarray, function(n, i){           return (n.type==="apple");         });         bananaarray  = $.grep(itemarray, function(n, i){           return (n.type==="banana");         });         kiwiarray = $.grep(itemarray, function(n, i){           return (n.type==="kiwi");         });                  itemarray=applearray.concat(kiwiarray).concat(bananarray);

Lochan Rao Pawar · Accepted Answer · 2011-11-24 14:20:55Z

0

use it this way

$(document).ready(function () {
  var data = {/*json-object*/};      
         ;
  data = data.sort(function (a, b) {
    if (a.type < b.type) { return -1 };
    if (a.type > b.type) { return 1 };
    return 0;
  });
  $.each(data, function (index, value) {
    $("#wrapper").append("<li>" + value.index + " : " + value.name + " * " + value.type + "</li>");
  });
});

answered Nov 24, 2011 at 14:20

Lochan Rao Pawar

233 bronze badges

Collectives™ on Stack Overflow

sorting an object respectively array effectively with jquery

2 Answers 2

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related