Jquery Random Array Delete

Question

var items = ["1","2","3","4","5","6","7","8","9","10","11","12"];
function bigger(){
    for(var i=0;i<items.length;i++){
        a = [Math.floor(Math.random() * items.length)+1];
        scale(a);   
    }   
}
function scale(number){
    $("#inner"+number+"").delay(100).transition({scale:1},300);
    items.splice(number,1);
} 
bigger();

here is my code ı try to delete numbers and use until finish array and i want to do it 1 by 1

Replace the for loop with while(items.length > 0), or your loop will stop before the array is empty. Remove the [] from a = [Math... - You're creating an array with one element and passing that array to scale(). — nnnnnn
– nnnnnn, Commented Jun 4, 2013 at 11:47
P.S. delay() only applies a delay to the animation queue on the individual element, so all of your elements will be processed at once. — nnnnnn
– nnnnnn, Commented Jun 4, 2013 at 11:53

adeneo · Accepted Answer · 2013-06-04 16:43:44Z

0

You'll need the index of the number in the array for splice(), not the string itself:

items.splice( items.indexOf(number), 1 );

It would probably be easier to just shuffle the array and pop of the last value to get a random array index, and at the same time removing it from the array:

var items = ["1","2","3","4","5","6","7","8","9","10","11","12"];

function scale(){
    var number = items.sort(function() { return 0.5 - Math.random();}).pop();
    $("#inner"+number+"").delay(100).animate({opacity:1},300, function() {
        scale();
    });
} 

scale();

not sure where the .transition() method comes from, but guessing it has a callback ?

FIDDLE

edited Jun 4, 2013 at 16:43

answered Jun 4, 2013 at 11:46

adeneo

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3 Comments

Uğur Oruc Over a year ago

thank you very much its work so fine and transition is ricostacruz.com/jquery.transit transform jquery easyly for every browser thnx work fine like what ı want

Uğur Oruc Over a year ago

everytime it gives same order i want to diffirent orders everytime :(

adeneo Over a year ago

@UğurOruc - My bad, seems I messed up the random function, fixed it now.

Arun P Johny · Accepted Answer · 2013-06-04 11:47:44Z

0

You need to

items.splice($.inArray(number, items),1);

answered Jun 4, 2013 at 11:47

Arun P Johny

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