Question

I got two DIV container each of it contains a puzzle (realization with jquery draggables).

I also got one object which looks like:

var puzzle var puzzle = {
    option1 = [],
    init:function(container) { .... }
}

To start the puzzle for one DIV i use:

// left puzzle div
var puzzleLeft = puzzle;
puzzleLeft.init(leftContainer);

// right puzzle div
var puzzleRight = puzzle;
puzzleRight.init(leftContainer);

But for the left one its not working. Everything I’m going to do left, happens in the right one :/ When I uncomment the start code for the right one, it will work for the left 🙂
Seems to be the right one start code is wrong?
Is this a wrong way to create 2 instances of the puzzle?

Thanks
Matt

Answer 1

You can create class’ constructor instead of object:

function Puzzle(){
   var self = this;

   self.option1 = [];
   self.init = function(container) { .... };
}

And create two instances of it:

// left puzzle div
var puzzleLeft = new Puzzle();
puzzleLeft.init(leftContainer);

// right puzzle div
var puzzleRight = new Puzzle();
puzzleRight.init(rightContainer);