Why is .join returning [object Object]

Question

Why is var test = positions.join("/"); returning [object Object]/[object Object]/[object Object] and so on? what needs to be changed in-order for this to work properly?

It should be returning positions like 0,0/0,360/0,660. Im not sure if the commas would be in there though.

$(function() {
    $('.AppList').droppable({
        accept: ".App",
        tolerance: 'fit',
        drop: function(event, ui) {
            var apps = $(".App"),
            positions = [];

            $.each(apps, function (index, app) {
                var positionInfo = $(app).position();

                positions.push(positionInfo);
            });
            var test = positions.join("/");
            console.log(test);
        }
    }); 
});

Because $(app).position() is returning an object like { top: 10, left: 10} — Catalin, Apr 27 '13 at 08:35

score 2 · Accepted Answer · answered Apr 27 '13 at 08:46

2

How about:

var test = JSON.stringify(positions);
console.log(test);

No need to invent your own serialization format.

answered Apr 27 '13 at 08:46

georg

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Yea this is working. Thank you :) @roasted Thanks for trying – ChristopherStrydom Apr 27 '13 at 08:50
How would i now split this back into an array? – ChristopherStrydom Apr 27 '13 at 08:51
@ChristopherStrydom: try `JSON.parse` – georg Apr 27 '13 at 08:52

A. Wolff · Answer 2 · 2013-04-27T08:46:17.413

1

You are returning object from function position(). Try this instead:

 $.each(apps, function (index, app) {
           var pos =  $(app).position(),   
               positionInfo = pos.top+","+ pos.left;

            positions.push(positionInfo);
        });

edited Apr 27 '13 at 08:46

answered Apr 27 '13 at 08:37

A. Wolff

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this is now returning [undefined Undefined]? – ChristopherStrydom Apr 27 '13 at 08:39
Use console.log($(app).position()) to see how looks your object, it seems not to be .X & .Y, maybe .top & .left – A. Wolff Apr 27 '13 at 08:40
okay it now looks like this Object { top=60, left=0 } – ChristopherStrydom Apr 27 '13 at 08:41
i need to use this `$.each(apps, function (index, app) {` because i have multiple divs and i want to get the positions of all of them. Here is a question i asked earlier http://stackoverflow.com/questions/16245137/array-slice-returning-object-object-instead-of-value – ChristopherStrydom Apr 27 '13 at 08:44
@ChristopherStrydom - Isn't that the very same problem, except with a different object. The solution is exactly the same. – Álvaro González Apr 27 '13 at 08:46
yea it is but i cant seem to find the solution even though you say its the same i don't know what syntax i need to use to make it work as before. – ChristopherStrydom Apr 27 '13 at 08:47

score 0 · Answer 3 · answered Apr 27 '13 at 08:38

0

The Array.join() method returns a string. Since your array items are objects you can't expect anything particularly useful—[object Object] is what JavaScript creates as default when casting objects to strings.

answered Apr 27 '13 at 08:38

Álvaro González

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Why is .join returning [object Object]

3 Answers3