You have to pass $super to the subclass' methods as its first argument
and then call it from there.
var Cat = Class.create(Animal, {
initialize: function($super, name) {
$super(name);
alert('Cat : initialize');
}
});
Best,
-Nicolas
On 9/20/07, Nicolás Sanguinetti <[EMAIL PROTECTED]> wrote:
> You have to pass $super to the subclass' methods as its first argument
> and then call it from there.
>
> var Cat = Class.create(Animal, {
> initialize: function($super, name) {
> $super(name);
> alert('Cat : initialize');
> }
> });
>
> Best,
> -Nicolas
>
> On 9/20/07, Les <[EMAIL PROTECTED]> wrote:
> >
> > I'm using Prototype 1.6 RC0. I don't see that the superclass
> > constructor is called automatically. They should always be called
> > automatically, and always before the subclass constructor. Do I have
> > to invoke the superclass constructor myself?
> >
> > See the code below:
> >
> > var Animal = Class.create({
> > initialize: function(name) {
> > alert('Animal: initialize');
> > this.name = name;
> > },
> > });
> >
> > // subclass that augments a method
> > var Cat = Class.create(Animal, {
> > initialize: function(name) {
> > alert('Cat : initialize');
> > },
> > });
> >
> > var cat = new Cat('test');
> >
> >
> > > >
> >
>
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