jQuery map() returns object instead of array

Question

I have the following code:

var selectedTitles = $("#js-otms-titles-table .select_title :checkbox:checked").map(function() {
    return $(this).attr('data-titleId');
});
console.log("Selected titles");
console.log(selectedTitles);

I expect that result will be an array. However I receive object like:

Object["55a930cd27daeaa6108b4f68", "55a930cd27daeaa6108b4f67"]

Is there a special flag to pass to the function? In docs they are talking about arrays as a return value. Did I miss something?

jQuery 1.11.2

charlietfl · Accepted Answer · 2017-05-10 00:05:36Z

41

$(selector).map() always returns jQuery object.

To get array from jQuery object use get()

var selectedTitles = $("#js-otms-titles-table .select_title :checkbox:checked").map(function() {
    return $(this).attr('data-titleId');
}).get();

edited May 10, 2017 at 0:05

answered Mar 1, 2016 at 20:36

charlietfl

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Comments

Scimonster · Accepted Answer · 2016-03-01 20:35:50Z

9

You need to call .get() on the final result. jQuery's .map() function returns a jQuery object (which can be convenient at times). .get() will fetch the underlying array that it's built on.

See http://api.jquery.com/map/

answered Mar 1, 2016 at 20:35

Scimonster

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2 Comments

Tamara Over a year ago

What is the difference between api.jquery.com/map and api.jquery.com/jquery.map ? I'm a little bit confused.

Scimonster Over a year ago

One works on a jQuery object (generally elements), the other on a plain object or array.

Bob Stein · Accepted Answer · 2021-08-18 18:00:54Z

7

The setup:

var a = array(1, 2, 3);
var f = function () { return process_each(this); };
var selector;   // e.g. "div.class" or $("div#id1, div#id2, div#id3")

Ways to map*:

$.map(a, f) - array in - array out

$(selector).map(f) - jQuery object in - jQuery object out

$(selector).map(f).get() - jQuery object in - array out (since jQuery 1.0)

$(selector).map(f).toArray() - jQuery object in - array out (since jQuery 1.4)

*In this context, to map means to process the items in your collection (array or jQuery object), one at a time with your function, and to output a collection (array or jQuery object) of the return values.

edited Aug 18, 2021 at 18:00

answered Nov 28, 2017 at 18:51

Bob Stein

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