dynamically loaded js function does not appear in Firebug js debugger

Question

There is a page1.html (I open it in browser):

<div id="content">
</div>
<script type="text/JavaScript">
    jQuery.ajax({
        type : "GET",
        url : 'page2.html',
        dataType : "html",
        success : function(response) {          
            jQuery('#content').append(response);

        }
    });
</script>

Code of the page2.html:

<script type="text/JavaScript" src="js/page2.js"></script>

<script type="text/JavaScript">
test();
</script>

Code of the page js/page2.js:

function test() {
    alert('Function Test()');
}

Everything works well, the window "Function Test()" is shown. But the problem is that I can't reach the code of function test() in firebug js debugger. It doesn't appear in event scripts nor in eval.

How can I fix that?

FYI: If I don't put the function in separate js file, but place it in page2.html, it appears in debugger corectly.

If I put "debugger" word in test() function, Firebug stops, but the source code of the function is still unreachable.

Versions: Firefox 3.0.10, Firebug 1.3.3

Update: pretty much the same as the question Making Firebug break inside dynamically loaded javascript, but there is no answer yet

score 4 · Answer 1 · answered May 17 '09 at 11:24

4

It's a hack, but I got firebug to stop in the external file js/page2.js by adding the word debugger twice. Once at the top of the file and another time at the top of the function.

If the debugger word is there only once in either place, firebug does not stop.

debugger;
function test() {
    debugger;
    alert('Function Test()');
}

answered May 17 '09 at 11:24

Jose Basilio

50,714
13
121
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Yes, it indeed stops on "debugger" if I place it anywhere in js/page2.js (I mentioned it in a message), but still I can't get the source code (watch local wars, step by step moving etc) – braz May 17 '09 at 12:33
By doing exactly what I mentioned in my post, having the debugger word twice, I am able to step through the source and watch local vars. – Jose Basilio May 17 '09 at 13:18
1

Thanks, it works indeed. However for more complex functions (like in my real case) it still doesn't work properly ... – braz May 17 '09 at 13:30
So, we can prove that it works in the simple scenario. Please provide a little more information on your complex case, and what you are trying to accomplish. Perhaps, there's a different way of approaching a solution. – Jose Basilio May 17 '09 at 17:37

score 0 · Answer 2 · edited Oct 10 '11 at 11:29

0

Try:

eval('debugger;');

It's not pretty but it seems to work.

edited Oct 10 '11 at 11:29

Jason Plank

2,336
5
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answered Mar 06 '11 at 12:29

Max Tardiveau

1

3

Why not just `debugger;`? `eval` seems unnecessary if you pass a static string. – Felix Kling Dec 17 '11 at 01:13

score 0 · Answer 3 · answered May 17 '09 at 11:01

0

What happens when you change js/page2.js to:

function test() {
    alert('Function Test()');
}
alert('loaded');

Do you see the "loaded" message?

answered May 17 '09 at 11:01

Phillip B Oldham

18,807
20
94
134

Yes, of course, Script execution works correctly, but it is not "loaded" in firebug properly – braz May 17 '09 at 12:34

dynamically loaded js function does not appear in Firebug js debugger

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