I always thought that firebug would send to the console content whenever the JavaScript encountered console.log('send this');

Then, I was messing around with jQuery's hover() method (something like $(".myDiv").hover(function () {...},function () {...console.log('leave');});, and couldn't figure out why an event was only happening once. Turns out it appears that firebug will only display the first occurrence of an identical console.log.

Then I added an identical console.log('leave'); after the first one, and not only did it display twice, but each time I hovered out, it displayed twice again.

Then, instead of duplicating the second console.log, I put a loop around it. This time it displayed only once, and wouldn't display on subsequent times I hovered out.

What is going on?

1 Answer 1


It is a new feature added in Firebug 1.12 called console grouping.

To disable it (requires version 1.12.1) :

  • Go to "about:config"
  • Promise you'll be careful
  • look for "extensions.firebug.console.groupLogMessages"
  • double-click on the option => the value is set to false

See also:



You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .