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Basic help using Firebug

Answer 1

A:

Yes you are using the console correctly, when i open firebug and type $("p") it returns everry p element in the DOM.

Ben Robinson 2010-07-20 15:32:00

Answer 2

A:

You're using the console correctly. Even if jQuery can't find any results, it should return an empty object, not null.

Could you console.log($); to see if jQuery is loaded?

That should result in the jQuery function being returned:

function ( selector, context ) {
    // The jQuery object is actually just the init constructor 'enhanced'
    return new jQuery.fn.init( selector, context );
}

kander 2010-07-20 15:32:41

Answer 3

+1 A:

Hi,

What you used is an ID selector. If you have a selector with an id u should use $("ID").

What you want is an array of css selectors-> then you should use $$("selector") -> in your case: $$("p")

more information is to be found here

http://www.joehewitt.com/software/firebug/docs.php

I hope this helped :D

Nealv 2010-07-20 15:42:41

Nealv - interesting! It appears both Firebug and jQuery use the dollar-function as a selector. It definitely explains why the topic starter was seeing the result he did, although the solution he was looking for was more regarding figuring out whether or not jQuery was loaded. Thanks for adding this answer, learned something today!

kander 2010-07-23 08:31:05

It was my understanding that he needed help with firebug. anyway glad to be of service. I learn lots here every day :)

Nealv 2010-07-23 11:07:20

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