Javascript that detects Firebug?

Original answer:

Check for the console object (created only with Firebug), like such:

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

Update (January 2012):

The Firebug developers have decided to remove window.console.firebug. You can still detect the presence of Firebug by duck typing like

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

or various other approaches like

if (document.getUserData('firebug-Token')) ...
if (console.log.toString().indexOf('apply') != -1) ...
if (typeof console.assert(1) == 'string') ...

but in general, there should be no need to actually do so.

If firebug is enabled, window.console will not be undefined. console.firebug will return the version number.

As of Firebug version 1.9.0, console.firebug is no longer defined because of privacy concerns; see release notes, bug report. This breaks the above mentioned methods. Indeed, it changes the answer to Allan's question to "there is no way"; if there is another way, it's considered a bug.

The solution instead is to check for the availability of console.log or whatever it is you want to use or replace.

Here is a suggestion for a replacement for the kind of code that David Brockman is presenting above, but one that doesn't remove any existing functions.

(function () {
    var names = ['log', 'debug', 'info', 'warn', 'error', 'assert', 'dir', 'dirxml', 
                'group', 'groupEnd', 'time', 'timeEnd', 'count', 'trace', 'profile', 'profileEnd'];

    if (window.console) {
        for (var i = 0; i < names.length; i++) {
            if (!window.console[names[i]]) {
                window.console[names[i]] = function() {};
            }
        }
    } else {
        window.console = {};
        for (var i = 0; i < names.length; i++) {
            window.console[names[i]] = function() {};
        }
    }
})();