Difference between $(array).each(...) and $.each(array, ...) in jQuery

Question

If I have an array in a JavaScript variable that I want to iterate through, I usually use jQuery's .each() function like this:

var myArray = ["foo", "bar"];
$(myArray).each(function(index, value) {
    console.log(value);
});

But I can achieve the same effect by passing in my array as an argument to the .each() function, like this:

var myArray = ["foo", "bar"];
$.each(myArray, function(index, value) {
    console.log(value);
});

Obligatory but unnecessary JSFiddle

This question can apply to basically any other jQuery function as well, not just .each(). Is there a functional difference between these two different usages?

Travis J · Accepted Answer · 2016-10-04 23:19:36Z

1

No, there is no difference in the way they function.

The main difference is going to be that using $.each() allows you to pass in an array whereas using $().each() requires you to have a constructed jQuery object to use as the context.

There is a slight difference in readability, but that is negligible.

If you were to look at jQuery's source code, you could also confirm this, as $.each is simply just a call to prototype's definition of each:

// Execute a callback for every element in the matched set.
each: function( callback ) {
    return jQuery.each( this, callback );
},

answered Oct 4, 2016 at 23:19

Travis J

82.5k43 gold badges213 silver badges280 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Difference between $(array).each(...) and $.each(array, ...) in jQuery

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related