Yes, this is a way to go:
var A = q.define({members : {a: function() {alert("a");}}});
var B = q.define({
extend: A,
members : {
a : function() {
this.constructor.base.prototype.a.call(this, arguments);
alert("b");
}
}
});
var b = new B();
b.a();
Am 26.09.2012 um 22:02 schrieb Loren Schlomer <[email protected]>:
> I guess this works:
>
> this.constructor.base.prototype.overriddenMethod(arguments, ...)
>
>
>
> On Wed, Sep 26, 2012 at 12:41 PM, Loren Schlomer
> <[email protected]> wrote:
>> One more question regarding q.define().
>>
>> Is it possible to invoke an overridden method in a super class?
>> this.base(arguments, ...) appears not to work.
>>
>>
>>
>>
>>
>> On Sun, Sep 23, 2012 at 11:25 PM, Martin Wittemann
>> <[email protected]> wrote:
>>> Hey,
>>>> ....does not work and emits the behavior described....And of course,
>>>> properties aren't supported with qx.Bootstrap.....
>>>>
>>> Thats right. :)
>>>
>>>> Slight overstep on my part, so indeed I was reading the docs incorrectly.
>>> Still, we need to enhance the docs about the define method. There are not
>>> enough samples and text to give a clear picture.
>>>
>>>> Should also note, you MUST have a members{} block defined (it can be
>>>> empty...) for q.define() to return a function prototype. Without it,
>>>> it returns the object I described yesterday. So a combination of
>>>> trying to use properties AND omitting members{} is what led to the
>>>> confusion.
>>>
>>> Sure, otherwise its defined as a static class and this does not need to be
>>> a function.
>>>
>>> Regards,
>>> Martin
>>>
>>>
>>>
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