更好地理解 Javascript OOP 架构

问题描述

当我阅读一些 Angularjs 的 UI 插件示例时，我偶然发现了一些代码，这些代码向我展示了我对 Javascript 的了解是相当有进步的:

As i read through some examples of Angularjs' UI add-on, i've stumbled over some code that showed me that my knowdledge of Javascript is quite improvable:

以下是 Angular 提供者内部的一个类:

The following is a class inside of an Angular provider:

function Dialog(opts) {

            var self = this, options = this.options = angular.extend({}, defaults, globalOptions, opts);
            this._open = false;

            this.backdropEl = createElement(options.backdropClass);
            if(options.backdropFade){
                // ...
            }

            this.handleLocationChange = function() {
                self.close();
            };

            // more functions
        }

非常简单.但是在该类之外，还有原型函数，例如上面调用的 close()

Pretty straightforward. But outside of that class, there are prototype functions, e.g the above invoked close()

Dialog.prototype.open = function(templateUrl, controller){
            var self = this, options = this.options;

            // .. some code
        };

现在我不明白为什么该函数被声明为原型，而是在类本身内部handleLocationChange.

Now i do not understand why that function is declared as a prototype, but handleLocationChange inside the class itself.

我如何决定选择哪种方法?

How do i decide which method to choose?

可以在这里找到完整的要点

推荐答案

考虑这两种情况:

Dialog.prototype.open = function...

Dialog.open = function....

第一种情况 - 每个通过调用 new Dialog() 创建的对象都会有这个 open 函数

First case - every object created by calling new Dialog() will have this open function

第二种情况与对话框对象无关，将其视为静态函数.

Second case has nothing to do with dialog objects, consider it as static function.

编辑

在这里找到了一个很好的答案:javascript-class-method-vs-class-原型方法

found a great answer here : javascript-class-method-vs-class-prototype-method

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