Javascript: calling function inside another function not working?

Question

So inside my <script> tags, I have these functions.

$.fn.getRandom = function(min, max) {
    return min + Math.floor(Math.random() * (max - min + 1));
}

function placeRandom() {
    alert(getRandom(1,7));
}

and I call the placeRandom() function like so

$(document).ready(function() {
    placeRandom();
});

when I do this, it gives me an

Object expected

error and refers to the line where I wrote

alert(getRandom(1.7));

Now, my main question is if there is anything wrong with the code which I am showing you, or does it look correct? Because if it looks correct then I think i'm going to have to run through the rest of my entire code and look for maybe a open bracket which isn't closed because that might be the problem. So yea, is there any problem with my placeRandom() or getRandom() function or is the error occuring because of another reason other than these functions?

You are binding getRandom to the jQuery namespace, you will only be able to access it through a jQuery object instance or via $.fn.getRandom() — Andrei Nemes
– Andrei Nemes, Commented Oct 16, 2013 at 14:51
with the way you are calling getRandom, i think you just want a standard function. getRandom function(min,max)... — PlantTheIdea
– PlantTheIdea, Commented Oct 16, 2013 at 14:51
Yes. That's exactly what I mean. Only attach a function to jQuery if you want to integrate it in the library (like a plug-in). See my answer — Andrei Nemes
– Andrei Nemes, Commented Oct 16, 2013 at 14:55

tleef · Accepted Answer · 2013-10-16 15:01:45Z

2

The problem is were you are defining the getRandom function. You are putting it in the jQuery.fn namespace. If you organize your code like the following it should work much better.

var getRandom = function(min, max) {
    return min + Math.floor(Math.random() * (max - min + 1));
}

var placeRandom = function() {
    alert(getRandom(1,7));
}

$(document).ready(function() {
    placeRandom();
});

Additional Info: jQuery.fn is actually a prototype but is commonly referred to as a namespace in this specific context.

edited Oct 16, 2013 at 15:01

answered Oct 16, 2013 at 14:52

tleef

3,6241 gold badge27 silver badges34 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

SLaks Over a year ago

That's a prototype, not a namespace.

tleef Over a year ago

Very true, I will clarify in my answer.

Andrei Nemes · Accepted Answer · 2013-10-16 14:52:50Z

1

You are binding getRandom to the jQuery namespace, you will only be able to access it through a jQuery object instance or via $.fn.getRandom()

So unless you want to be able to call it like $(selector).getRandom(), just declare it as a standard function.

answered Oct 16, 2013 at 14:52

Andrei Nemes

3,1221 gold badge18 silver badges23 bronze badges

4 Comments

SLaks Over a year ago

That's a prototype, not a namespace.

Andrei Nemes Over a year ago

It's a namespace pointing to the jQuery prototype. Wanna argue semantics?

Andrei Nemes Over a year ago

Yes, it will. Try it :)

Andrei Nemes Over a year ago

In JavaScript, functions are first class citizens, which means you can do a lot of fun stuff with them, such as passing them as arguments to other functions, storing their references in variables or using them in objects.

mador · Accepted Answer · 2013-10-16 14:54:19Z

1

you are binding getRandom to the $ namespace (jQuery). so you have to call it $.fn.getRandom(1,7) or define the function the following:

function getRandom(min, max) {
    return min + Math.floor(Math.random() * (max - min + 1));
}

answered Oct 16, 2013 at 14:54

mador

1,2239 silver badges13 bronze badges

Collectives™ on Stack Overflow

Javascript: calling function inside another function not working?

3 Answers 3

2 Comments

4 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

4 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related