How do I print debug messages in the Google Chrome JavaScript Console?

04/17/2020 23:00:02

How do I print debug messages in the Google Chrome JavaScript Console?

Please note that the JavaScript Console is not the same as the JavaScript Debugger; they have different syntaxes AFAIK, so the print command in JavaScript Debugger will not work here. In the JavaScript Console, print() will send the parameter to the printer.

Verified Answer (597 Votes)

10/20/2008 16:20:23

Executing following code from the browser address bar:

javascript: console.log(2);

successfully prints message to the "JavaScript Console" in Google Chrome.


Answer #2 (166 Votes)

05/03/2010 16:01:31

Improving on Andru's idea, you can write a script which creates console functions if they don't exist:

if (!window.console) console = {};
console.log = console.log || function(){};
console.warn = console.warn || function(){};
console.error = console.error || function(){}; = || function(){};

Then, use any of the following:


These functions will log different types of items (which can be filtered based on log, info, error or warn) and will not cause errors when console is not available. These functions will work in Firebug and Chrome consoles.


Answer #3 (47 Votes)

10/22/2013 12:44:26

Just add a cool feature which a lot of developers miss:

console.log("this is %o, event is %o, host is %s", this, e,;

This is the magical %o dump clickable and deep-browsable content of a JavaScript object. %s was shown just for a record.

Also this is cool too:

console.log("%s", new Error().stack);

Which gives a Java-like stack trace to the point of the new Error() invocation (including path to file and line number!).

Both %o and new Error().stack are available in Chrome and Firefox!

Also for stack traces in Firefox use:


As says.

Happy hacking!

UPDATE: Some libraries are written by bad people which redefine the console object for their own purposes. To restore the original browser console after loading library, use:

delete console.log;
delete console.warn;

See Stack Overflow question Restoring console.log().

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