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Why is Firefox 3 breaking with console.log

I have the following:

console.log (a.time_ago() + ' ' + b.time_ago());

This is breaking in FireFox 3, meaning when FF hits that line in the JS, it goes no further. Strangely if I have Firebug open it doesn't break and continues as normal. Some how firebug prevents this issue?

I'm puzzled on this one. Any thoughts as to why console.log would break firefox 3, but not if firebug is open?

Thanks

like image 704
AnApprentice Avatar asked Feb 10 '11 22:02

AnApprentice


3 Answers

This is not just Firefox. Your code will stop working in every browser (except Chrome and safari (in some instances) because they have console.log() built in along with their developer tools.)

It is because when you don't have firebug open, the object "console" is not defined. You should take care never too leave console.log() functions in your code, or it will break in every browser.


I'd like to add that I have sometimes used this function:

function log () {
    if (typeof console == 'undefined') {
        return;
    }
    console.log.apply(console, arguments);
}

Then you can simply call:

log(somevar, anothervar);

and it will work the same way as console.log, but will not fail if firebug is not loaded (and is shorter to type :P)

Cheers

like image 166
arnorhs Avatar answered Nov 10 '22 08:11

arnorhs


In case, that firebug is closed, I overwrite the console object. So, you can implement fallback functions ...

console = console || { log : function() {
// place your logging code here, if firebug is closed
}, debug : function() {
// place your debug code here, if firebug is closed
} /*, [ and so on .. ] */ };

Greetings,

Dyvor

like image 42
BaggersIO Avatar answered Nov 10 '22 10:11

BaggersIO


In FireFox if the console isn't open when you call to it a JavaScript error is thrown.

I wrap all my console.log's in a wrapper to check for console - you can either wrap the check around the console call or use a different name to alias console.

Aliasing

/* konsole is a safe wrapper for the Firebug console. */
var konsole = {
  log: function(args){},
  dir: function(args){},
  debug: function(args){},
  info: function(args){},
  warn: function(args){},
  error: function(args){}
};
// Remove below here when in production
if (typeof window.console != 'undefined' && typeof window.console.log == 'function') {
  konsole = window.console;
}
konsole.log('testing debugging');
konsole.error('throw an error message to the console');

Check for console

if (typeof window.console != 'undefined' && typeof window.console.log == 'function') {
  console.log('testing debugging');
  console.error('throw an error message to the console');
}
like image 2
Terri Ann Avatar answered Nov 10 '22 09:11

Terri Ann