Getting element `index` out of the `$.initialize` loop.

[Irving Armenta]

Hello, I'm trying to get the index number of the elements out in the $.initialize loop, like this:

$.initialize('.element', function(i,el) { 
  console.log(i) // this always return: 0 
});

What I want to achieve, is the same functionality as $.each or $.map to count and distinguish the elements in the DOM, but I keep getting 0 on index, however the DOM nodes are shown correctly on el.

Is there a solution for this?

[Adam Pietrasiak]

I think concept of element index is not relevant here as it's calling this callback as soon as element has appeared on the screen, so it has no context of 'group' like map or each has.

Consider such case

Initially you have 5 .element divs
Then you add 2 more .element divs
Then you remove 3 of existing .element divs
Then you add one more ----> what would you expect now to be i inside this function?

[Irving Armenta]

I will expect the same result I get using $.each or $.map , to get the actual length of the elements that are exactly at that time in the DOM, in your example:

5 .element divs (0,1,2,3,4)
add 2 = 7 .element divs (0,1,2,3,4,5,6)
remove 3 = 4 .element divs (0,1,2,3)
add 1 = .element divs (0,1,2,3,4)

You can do this by setting a function with $.each and run a function to "read" again the elements in the DOM, but that is exactly what I want to $.initialize to do for me, I don't want to set up a function to update the list of elements.

[Damien Bezborodov]

An iterator doesn't exactly make sense in this context since elements can be inserted into the DOM out of order, and also can appear in complex or nested trees, not just as siblings.

If you are insistent on this, you can off course do:

var i = 0;
$.initialize('.element', function(el) { 
  console.log(i);
  i++;
});

Although I understand that it may make sense to have an iterator in some circumstances, nobody has asked for such a feature to date. I wonder for the sake of curiosity what your scenario is that you need this for?

[Damien Bezborodov]

Wontfix.