How to quickly and conveniently disable all console.log statements in my code?

Question

Is there any way to turn off all console.log statements in my JavaScript code, for testing purposes?

Thanks!

Answer 1

As far as I can tell from the documentation, Firebug doesn't supply any variable to toggle debug state. Instead, wrap console.log() in a wrapper that conditionally calls it, i.e.:

DEBUG = true; // set to false to disable debugging
function debug_log() {
    if ( DEBUG ) {
        console.log.apply(this, arguments);
    }
}

To not have to change all the existing calls, you can use this instead:

DEBUG = true; // set to false to disable debugging
old_console_log = console.log;
console.log = function() {
    if ( DEBUG ) {
        old_console_log.apply(this, arguments);
    }
}

Answer 2

The following is more thorough:

var DEBUG = false;
if(!DEBUG){
    if(!window.console) window.console = {};
    var methods = ["log", "debug", "warn", "info"];
    for(var i=0;i<methods.length;i++){
     console[methods[i]] = function(){};
    }
}

This will zero out the common methods in the console if it exists, and they can be called without error and virtually no performance overhead. In the case of a browser like IE6 with no console, the dummy methods will be created to prevent errors. Of course there are many more functions in Firebug, like trace, profile, time, etc. They can be added to the list if you use them in your code.

You can also check if the debugger has those special methods or not (ie, IE) and zero out the ones it does not support:

if(window.console && !console.dir){
var methods = ["dir", "dirxml", "trace", "profile"]; //etc etc
    for(var i=0;i<methods.length;i++){
     console[methods[i]] = function(){};
    }
}

Answer 3

I'm havin a problem with the console error, were is this progrm so I can just remove it. Thanks