Accessing parent class in Backbone

Question

I need to call the initialize method of the parent class, from inside the inherited MyModel-class, instead of completely overwriting it as I am doing today.

How could I do this?

Here's what my code looks right now:

BaseModel = Backbone.Model.extend({
    initialize: function(attributes, options) {
        // Do parent stuff stuff
    }
});

MyModel = BaseModel.extend({
    initialize: function() {
        // Invoke BaseModel.initialize();
        // Continue doing specific stuff for this child-class.
    },
});

Solution 1

MyModel = BaseModel.extend({
    initialize: function() {
        MyModel.__super__.initialize.apply(this, arguments);
        // Continue doing specific stuff for this child-class.
    },
});

Solution 2

Try

MyModel = BaseModel.extend({
    initialize: function() {
        BaseModel.prototype.initialize.apply(this, arguments);
        // Continue doing specific stuff for this child-class.
    },
});

Solution 3

This worked for me, when I was trying to inherit among my models:

MyModel.prototype.initialize.call(this, options);

Referenced from http://documentcloud.github.com/backbone/#Model-extend

Thanks.

Solution 4

I think it'd be

MyModel = BaseModel.extend({
    initialize: function() {
        this.constructor.__super__.initialize.call(this);
        // Continue doing specific stuff for this child-class.
    },
});

Solution 5

this seems to be almost a duplicate of Super in Backbone, so you want something like this:

Backbone.Model.prototype.initialize.call(this);

View more solutions

Comments

[Raynos]

It's this.__super__.initialize.call(this);

[Sander]

i doubt rewriting backbone is in order here :) we got to manage with the structure backbone gives us, and there are far better solutions than rewriting all this, like the super is more than enough.

[Industrial]

Both this.__super__.prototype.initialize.call(this); and this.__super__.initialize.call(this); throws this.__super__ is undefined in Firebug.

[Raynos]

@Industrial then it's doing something silly. Try this.__super__.initialize.apply(this, arguments);

[Yury Tarabanko]

__ super __ is not intended to use it directly, as the underscored name implies.

[wheresrhys]

Try this.constructor.__super__.initialize.call(this);

[Raynos]

@YuryTarabanko sure it is, it's just someone choose a __super__ naming convention in their inherits utility function to avoid naming conflicts. It's exposed by backbone

[Yury Tarabanko]

@Raynos: Why not BaseModel.prototype.initialize.apply(this, arguments);? this should work without __super__.

[Raynos]

@YuryTarabanko either works. The __super__ solution is more elegant because it doesn't require changing when you rename BaseModel.

[Yury Tarabanko]

documentcloud.github.com/backbone/#Model-extend Brief aside on super: JavaScript does not provide a simple way to call super — the function of the same name defined higher on the prototype chain. If you override a core function like set, or save, and you want to invoke the parent object's implementation, you'll have to explicitly call it...

[Salman von Abbas]

In Ember.js <3 you just do this._super(). tada! it's magic. :)

[Raynos]

@SalmanPK I'm sure that will break horribly in plenty of edgecases. It's impossible to implement a sensibly semantic super in JS.

[Scott Weaver]

this appears to work even in the presence of _.bindAll(this) calls, which render the __super__ property of the constructor undefined.

[Khalid Dabjan]

I wanted to pass arguments to the initialize function, and this is the only version that worked for me this.constructor.__super__.initialize.call(this,arguments);

[MikeSchinkel]

FWIW Jeremy Ashkenas states that __super__ is not intended to be used directly.

[MikeSchinkel]

FWIW Jeremy Ashkenas states that __super__ is not intended to be used directly.

[MikeSchinkel]

FWIW Jeremy Ashkenas states that __super__ is not intended to be used directly.

[Marcel Falliere]

better than using __super__ has the underscores suggests a private member