I can't access array/obj outside jQuery event function

Question

So I've got an object, or array, declared at the beginning of anything, outside everything:

var Thing = {title:'horse'};

Then I've got:-

$('.clickedIt').fadeOut(200, function() { console.log(Thing.title); }

That will fail. However, if I place above that same console log out of fadeOut, it'll be fine.

How does it fail? Where is the fadeOut called with respect to Thing? — pimvdb
– pimvdb, Commented Aug 17, 2012 at 15:54
can you provide a more complete script for us to look at? The problem here will laregely depend on the scope of Thing. — jackwanders
– jackwanders, Commented Aug 17, 2012 at 15:55

Paul · Accepted Answer · 2012-08-17 16:01:22Z

4

If you want something to be global, just define it on the window Object.

window.Thing = { title: 'horse '};

Then use it like so:

$( '.clickedIt' ).fadeOut(200, function() { 
    console.log( window.Thing.title ); 
});

Just a note, putting a number of variables on the window Object is not recommended, I would recommend looking into name-spacing: http://addyosmani.com/blog/essential-js-namespacing/

Here is an example:

//simple JavaScript module
( function( window ) {
    //define your applications root namespace
    window.myApp = {
        Thing: { title: 'horse '}
    };
})( window );

//jQuery ready function
$( function() {
    $( '.clickedIt' ).fadeOut( 200, function() { 
        console.log( myApp.Thing.title ); 
    });
});

edited Aug 17, 2012 at 16:01

answered Aug 17, 2012 at 15:55

Paul

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Comments

Adil · Accepted Answer · 2012-08-17 15:58:56Z

1

The value you assigned to title horse is undefined wrap it within quotes to make it string litral,

Live Demo

var Thing = {title:'horse'};
$('.clickedIt').fadeOut(200, function() { console.log(Thing.title); })

answered Aug 17, 2012 at 15:58

Adil

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5 Comments

pimvdb Over a year ago

I didn't downvote, but the OP stated that it does work with the console.log put outside, so horse is not the issue.

waxical Over a year ago

It appeared as though you had just copy and paste the question. I see your alteration now you explain, but that was a typed demo, so it's not the issue.

Adil Over a year ago

I checked on fiddle it gave me error for horse undefined @pimvdb

waxical Over a year ago

Yes, you solved another question, understandably... hence the down vote was reversed. Thanks.

Adil Over a year ago

I solved this question check the accepted answer that also wrapped horse within quotes. I could not exactly understand what you mean?

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