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  • 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

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