views:

1391

answers:

2

Hello,

I am trying to achieve something like the following but dont know whats wrong:

$.a = function() {

// some logic here

function abc(id) {
   alert('test'+id);
}


}

$.a.abc('1');

I tried using the return function, but that doesnt seem to work either. Can someone please help.

Thank you for your time.

+6  A: 

Since $.a must be a function in itself, you'll have to add the abc function as a property to the $.a function:

$.a = function () {
    // some logic here...
};

$.a.abc = function (id) {
    alert('test' + id);
};

If abc must be defined from within the $.a function, you can do the following. Do note that $.a.abc will not be available until $.a has been called when using this method! Nothing inside a function is evaluated until a function is called.

$.a = function () {

    // Do some logic here...

    // Add abc as a property to the currently calling function ($.a)
    arguments.callee.abc = function (id) {
        alert('test' + id);
    };
};

$.a();
$.a.abc('1');
Blixt
but i want it to be a function within a function. i cannot change the structure of $.a.
Alec Smart
Ah, I see. I'll update my example.
Blixt
Cool, nice answer! Thank you.
John McCollum
And just because I've never seen it before, here's a nice write up of what arguments.callee does: https://developer.mozilla.org/En/Core_JavaScript_1.5_Reference/Functions_and_function_scope/arguments/callee
John McCollum
A: 
$.a = (function(){
    var a = function() {
        //...
    };
    a.abc = function() {
        //...
    }
    return a;
})();
J-P