Javascript two-dimensional array

Question

I want to create a two dimensional array in a javascript function. I found code that should do that but doesn't. I declare the array then define a function to add elements to the array which are also arrays.

 // Array function
 var card_array = new Array();
 function card_array(card_id, card_top, card_left) {
 alert('array');
 this.card_id = card_id;
 this.card_top = card_top;
 this.card_left = card_left;
 }
 // Toggle LinkCard minimize/expand
 function toggle_linkcard(toggle, card_id) {
 var icard = 0; 
 $('.linkcard').each(function () {
 card_top = $(this).position().top;
 card_left = $(this).position().left;
 card_i = $(this).attr('id');
 card_array[card_array.length++] = new card_array(card_i, card_top, card_left);
 icard++;
 });
 alert(card_array);
 }

The line of code where I add elements to the array breaks the code.

 card_array[card_array.length++] = new card_array(card_i, card_top, card_left);

What should I fix in that?

Answer 1

You defined the function's name as card_array, same name as the variable's. So after that line of code, you don't have any variable named card_array, only the function. Try changing your variable or function name.

• Jonathan M: Also, you can eliminate some code clutter by doing a card_array.push() instead of card_array[card_array.length++].

Answer 2

The problem here is that you have two values with the same name: card_array

• A variable named which is initialized to a new Array()
• A function which takes 3 parameters

The function declaration happens last and hence wins. So when you execute the expression card_array[card_array.length++] you are doing so on a function instance, not an array.

To fix this change the function name to a unique name.

• user823527: That was the problem. Thanks.

Answer 3

Just change this line:

var card_array = new Array();

into:

var my_card_array = new Array();

And this one:

card_array[card_array.length++] = new card_array(card_i, card_top, card_left);

into:

my_card_array.push(new card_array(card_i, card_top, card_left));

And of course, change the alert.

• user823527: Thanks. I changed the variable name and switched to .push(). This works.