Just starting using the Firebug console. I have a test script which I'll post below. I open up the Firebug console and type $("p"); this returns null. Its my understanding it should return all my p elements ie p, p.foo, p, p#bar. A conflict maybe or am I just using the console incorrectly?
<!DOCTYPE html>
<html>
<head>
<title>Testing jQuery</title>
</head>
<body>
<p>Hello World!</p>
<p class="foo">Another paragraph, but this one has a class.</p>
<p><span>This is a span inside a paragraph.</span></p>
<p id="bar">Paragraph with an id.
<span class="foo">And this sentence is in a span.</span>
</p>
<script type="text/javascript" src="http://www.google.com/jsapi"></script>
<script type="text/javascript"
google.load("jquery", "1.4.2");
</script>
</body>
</html>
3 Answers 3
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
2 Comments
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 );
}
Comments
Yes you are using the console correctly, when i open firebug and type $("p") it returns everry p element in the DOM.
>in your script that is loading jquery, or is that a typo?google.load("jquery", "1.4.2");... is that a copy and paste error?