You can wrap $() around an array of DOM elements, or around a jQuery object, but not around an array of jQuery objects

Question

Not really sure how to phrase this question so it's generic enough. (and will rephrase, after I know what I'm asking). The problem, I believe deals with variables in a JQuery .find() function.

Problem: you can wrap $() around an array of DOM elements, or around a jQuery object, but not around an array of jQuery objects

The best I can do, at this time, is provide an example here

\n

The problem code in the previous fiddle, is here:

////////////////Neither of the following works////////////////
//nodelevel = nodesWithMinuses.find('div.node.level' + levelnumber);
  nodelevel = $(nodesWithMinuses).find('div.node.level' + levelnumber);
////////////////Neither of the previous works////////////////

Answer 1

Apparently you can wrap $() around an array of DOM elements, or around a jQuery object, but not around an array of jQuery objects.

Try changing this line:

var nodeWithMinus = thisplusminus.parent().parent();

to this:

var nodeWithMinus = thisplusminus.parent().parent()[0];

This will extract the DOM element and turn nodesWithMinuses into an array of DOM elements.

Answer 2

The variable nodesWithMinuses is an array of jQuery objects. You cannot apply jQuery method to anything else then a jQuery object. You have 2 options : either you declare nodesWithMinuses as a jQuery object and add objects to it via the add method :

var nodesWithMinuses = $();
//...
nodesWithMinuses.add(element);
// instead of nodesWithMinuses.push(element);

either find a way to convert the array to a jQuery object :

$($.map(nodesWithMinuses,function(el,i){return el.get(0)})).find('div.node.level' + levelnumber);

or traverse the array and search for the element in each object at a time :

var result = $();
$.each(nodesWithMinuses,function(i,el){
    result.add(nodesWithMinuses.find('div.node.level' + levelnumber));
});
console.log(result);

Answer 3

In your code, nodeWithMinuses is an array but you're trying to wrap it in the jQuery function. That doesn't work on arrays. Using a variable in jQuery's find function is fine.