jQuery plugin .fn question
- by user319342
Hello,
I seem to be having some problems creating a jquery plugin. I am testing a simple plugin in firebug console and the function is not being defined as I would expect. Here is the code pattern I am using
jQuery.fn.test = function () {console.log("runs")}
I am trying to call the function with this call which is not working.
$.test()
But this call does
$.fn.test()
I don't think this is how it is supposed to work so I think I am doing something wrong, although all the documentation seems to agree that I should be correct. Any one have any advice?
Thanks,
Cory