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Home/ Questions/Q 7896831
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Editorial Team
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Editorial Team
Asked: June 3, 20262026-06-03T07:59:42+00:00 2026-06-03T07:59:42+00:00

I am ciruious to know what are the reasons that could cause a function

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I am ciruious to know what are the reasons that could cause a function “not being” one.
For example, I have the following:

$.fn.myfunction = function(options){
  alert("test");
};
  $("#hello").myfunction();
alert($.fn.myfunction instanceof Function);

Why would FireBug, log that it is not a function?

Thanks in advance.

Edit:
I would like a list of all the possibilities that could be the reason for the error.
This isn’t an error I got, but I just want to widen my perspective and be aware of more possibilities.

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  1. Editorial Team
    Editorial Team
    2026-06-03T07:59:43+00:00Added an answer on June 3, 2026 at 7:59 am

    Setting $.fn.myfunction makes the myfunction function available to the object returned by $(selector), not to $ itself.

    Thus, $("#hello").myfunction is a function but $.myfunction is not. If you really want $.myfunction to be a function for some reason (e.g., it’s a utility function that doesn’t need a jQuery object list to operate), just set it explicitly, without using $.fn:

    $.myfunction = function() { .... }
    
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