CODEWITHSUNDEEP
javascriptfirefoxfirebug
Matt Brailsford
Matt Brailsford

Reputation: 2237

Detect firebug opening and closing

Does anybody have a way to detect Firebug opening and closing.

I know you can do the following:

if (window.console && window.console.firebug) {
  //Firebug is enabled
}

but this only detects the firebug console on page load. What I want to do is on a page where firebug is not open, detect the opening of the firebug console.

I've tried the following, but with no luck

setInterval(function(){
  if(window.console && window.console.firebug){
    ...
  else
    ...
}, 1000);

Any help greatly appreciated.

Matt

Upvotes: 2

Views: 2845

Answers (6)

Eduardo Gamero Bengoa
Eduardo Gamero Bengoa

Reputation: 1

If you only want to detect Firebug opening/closing you can check window resize/blur events.

Upvotes: 0

Mohammed Shannaq
Mohammed Shannaq

Reputation: 835

you can make it as

var t = setTimeout("CheckFireBug()",10000);
function CheckFireBug() {
    var t = setTimeout("CheckFireBug()",10000);
    if (window.console && window.console.firebug) {
        //Firebug is enabled
        console.debug('Firebug is enabled.');
    } else {
        //Firebug is not enabled
        console.debug('Firebug is not enabled.');
    }
}

Upvotes: 1

cuixiping
cuixiping

Reputation: 25381

console.table() returns "_firebugIgnore" if firebug is running.

if( window.console && (console.firebug || 
   console.table && /firebug/i.test(console.table()) ))
{
    alert('Firebug is running');
}else{
    alert('Firebug is not found');
}

Upvotes: 1

user123444555621
user123444555621

Reputation: 152996

Firebug overwrites the console property of window, so you may detect it like so:

var _console = window.console;
Object.defineProperty(window, 'console', {
    set: function (x) {
        if (x.exception) { // check if this is actually Firebug, and not the built-in console
            alert('Firebug on');
        }
        _console = x;
    },
    get: function () {
        return _console;
    }
});

The problem is that this object remains when Firebug is closed, so you can't detect that. Maybe there's some other way but I can't find it ATM.

Details: It's not possible to access Firebug's execution context from document scripts, so we're limited to waiting for Firebug to access some of window's properties, which does not seem to happen when you close Firebug. Here's some of the events during shutdown, taken with FBTrace:

Firebug stack trace

I've searched the stacktrace for "leaks" to window, but couldn't find any.

Upvotes: 1

johnjbarton
johnjbarton

Reputation: 1857

The Firebug window.console object is created just before the first Javascript in the page is executed but only if Firebug is active for the page before the first JS and if the user has the Firebug Console panel enabled.

In other words, from within the page you can only detect if the Console is enabled. But for your purposes that should be enough.

We should delete the console property if a user turns Firebug off for a page. I don't know if we actually do that.

Upvotes: 2

BigbangO
BigbangO

Reputation: 820

Simply.. You cant. The firebug window not just an another couple of div element on your page.

Upvotes: 2

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