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问题:
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.
},
});
回答1:
MyModel = BaseModel.extend({
initialize: function() {
MyModel.__super__.initialize.apply(this, arguments);
// Continue doing specific stuff for this child-class.
},
});
回答2:
Try
MyModel = BaseModel.extend({
initialize: function() {
BaseModel.prototype.initialize.apply(this, arguments);
// Continue doing specific stuff for this child-class.
},
});
回答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.
回答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.
},
});
回答5:
this seems to be almost a duplicate of Super in Backbone, so you want something like this:
Backbone.Model.prototype.initialize.call(this);
回答6:
Similar to @wheresrhys, but I would use apply instead of call in case BaseModel.initialize is expecting arguments. I try to avoid processing the attributes map that can be passed to a Backbone Model upon initialization, but if the BaseModel were actually a View or a Collection then I might want to set options.
var MyModel = BaseModel.extend({
initialize: function() {
this.constructor.__super__.initialize.apply(this, arguments);
// Continue doing specific stuff for this child-class.
},
});
回答7:
here's a multi generation callSuper method, just add it to your extending class.
callSuper: function (methodName) {
var previousSuperPrototype, fn, ret;
if (this.currentSuperPrototype) {
previousSuperPrototype = this.currentSuperPrototype;
// Up we go
this.currentSuperPrototype = this.currentSuperPrototype.constructor.__super__;
} else {
// First level, just to to the parent
this.currentSuperPrototype = this.constructor.__super__;
previousSuperPrototype = null;
}
fn = this.currentSuperPrototype[methodName];
ret = (arguments.length > 1) ? fn.apply(this, Array.prototype.slice.call(arguments, 1)) : fn.call(this);
this.currentSuperPrototype = previousSuperPrototype;
return ret;
}
回答8:
You might consider rewriting your code using functional inheritance.
var BackBone=function(){
var that={};
that.m1=function(){
};
return that;
};
var MyModel=function(){
var that=BackBone();
var original_m1=that.m1;
//overriding of m1
that.m1=function(){
//call original m1
original_m1();
//custom code for m1
};
};