Javascript Prototype Programming

Question

I'm trying to include some namespace into my Javascript API.

Here's what I have so far:

if (!Sample || Sample == "undefined")
    var Sample = {};

Sample.SomeApi = {};

Sample.SomeApi.prototype = {
    SomeMethod: function() {
        alert('some api');
    }
};

What's going on here?

When I'm calling Sample.SomeApi.SomeMethod(); // it won't work as it will complain:

Uncaught TypeError: Object #<Object> has no method 'SomeMethod'
(anonymous function)Course:43
onclick

Side note: you want undefined or typeof Sample === 'undefined' not "undefined". — benastan
– benastan, Commented Jan 10, 2012 at 5:52
Well, Sample == "undefined" is not going to test if it is undefined, it is going to test whether Sample equals the string "undefined." — benastan
– benastan, Commented Jan 10, 2012 at 6:03
Good point, alternatively, he can do Sample == undefined but then again the whole undefined is unnecessary, that's what !Sample is testing for anyway. — Luis Perez
– Luis Perez, Commented Jan 10, 2012 at 6:33

Luis Perez · Accepted Answer · 2012-01-10 06:28:18Z

2

I think you meant __proto__ not prototype, like so:

if (!Sample || Sample == "undefined")
    var Sample = {};

Sample.SomeApi = {};

Sample.SomeApi.__proto__ = {
    SomeMethod: function() {
        alert('some api');
    }
};

Sample.SomeApi.SomeMethod();

You can test it out in this jsFiddle: http://jsfiddle.net/luisperezphd/xtyyf/

But why not just this:

if (!Sample || Sample == "undefined")
    var Sample = {};

Sample.SomeApi = {
    SomeMethod: function() {
        alert('some api');
    }
};

Sample.SomeApi.SomeMethod();

It's much cleaner.

And there is the jsFiddle for that: http://jsfiddle.net/luisperezphd/HUuMQ/

edited Jan 10, 2012 at 6:28

answered Jan 10, 2012 at 5:48

Luis Perez

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2 Comments

Daff Over a year ago

I'm not so sure about that. That would mean that you assign SomeMethod to the prototype of {}?

clockworkgeek Over a year ago

Using __proto__ works because objects do not have a prototype, functions do. However __proto__ is not standard and won't work across browsers, that is what getPrototypeOf is for. But the second method really is much cleaner.

Daff · Accepted Answer · 2012-01-10 05:45:52Z

1

You have to create a constructor for Sample.SomeApi not assign an empty object:

if(typeof Sample == 'undefined')
    Sample = {};

Sample.SomeApi = function() {
};

Sample.SomeApi.prototype = {
    someMethod : function() {
            alert('some api');
    }
}

new Sample.SomeApi().someMethod();

answered Jan 10, 2012 at 5:45

Daff

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2 Comments

LB. Over a year ago

how would you make it call it like this: Sample.SomeApi.someMethod(); instead of Sample.SomeApi().someMethod(); ?

Daff Over a year ago

Well you don't need to use the prototype if your API is only a single object. just use Sample.SomeApi = { SomeMethod: function() { alert('some api'); } );

AmGates · Accepted Answer · 2012-01-10 06:41:58Z

0

Try this

if(typeof Sample == 'undefined')
        var Sample ={}
    Sample.SomeApi = function(a)
    {
        this.a =a;
    }

Sample.SomeApi.prototype.SomeMethod = function() {
        alert('some api');
    }
    var d = new Sample.SomeApi(1);
    d.SomeMethod();

Hope this helps you

Go through this link here

answered Jan 10, 2012 at 6:41

AmGates

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