Difference between console.log() and console.debug()?

Question

Technically console.log console.debug and console.info are identical However the way they display the data is little different. console.debug is not visible by default in the browser's JS console. It can be enabled by using the console's filter options.

console.log Black color text with no icon

console.info Blue color text with icon

console.debug Pure black color text

console.warn Yellow color text with icon

console.error Red Color text with icon

var playerOne = 120;
var playerTwo = 130;
var playerThree = 140;
var playerFour = 150;
var playerFive = 160;

console.log("Console.log" + " " +  playerOne);
console.debug("Console.debug" + " " +playerTwo);
console.warn("Console.warn" + " " + playerThree);
console.info("Console.info" + " " + playerFour);
console.error("Console.error" + " " + playerFive);

For at least IE, Firefox and Chrome consoles, .debug() is just an alias for .log() added for improved compatibility

https://developer.mozilla.org/en-US/docs/Web/API/console

https://developers.google.com/chrome-developer-tools/docs/console-api#consoledebugobject_object

https://msdn.microsoft.com/en-us/library/ie/hh772183(v=vs.85).aspx

They are almost identical - the only difference is that debug messages are hidden by default in recent versions of Chrome (you have to set the log level to Verbose in the Devtools topbar while in console to see debug messages; log messages are visible by default).

console.info ,console.debug methods are identical to console.log.

• console.log Printing statement
• console.info Black color text with "i" icon in blue color
• console.debug Blue Color text

Documentation:

• Chrome
• IE
• Edge
• FireFox
• Safari

If you want the ability to disable logging after a product is finished you could override the console.debug() function or make another custom one.

console.debug = function() {
    if(!console.debugging) return;
    console.log.apply(this, arguments);
};

console.debugging = true;
console.debug('Foo', {age:41, name:'Jhon Doe'});

Foo▸ {age: 41, name: "Jhon Doe"}

console.debugging = false;
console.debug('Foo', {age:26, name:'Jane Doe'});

No output

However I havent figured a way to color the outputs as well.