Question

var myclass = {
    init:function () {
        this.customer = null;
    },
    test : function(data){
      alert(testing);
    }
};

我正在像上面那样实例化myclass，后来我试图调用类的方法test，但它不起作用。我做错了什么？

var testClass = new myclass.init();
testClass.customer = 'John B';
testClass.test(); //doesnt alert 1

由于某种原因，我没有收到警报，而是收到此错误：

未捕获TypeError：对象[object Object]没有方法'test'

Answer 1

您必须将“类”定义为构造函数，而不是对象文字：

var MyClass = function(){
    this.init = function () {
        this.customer = null;
    };

    this.test = function(data){
      alert('testing');
    };
};
var testClass = new MyClass();
testClass.init();
testClass.customer = 'John B';
testClass.test(); //alerts 'testing'

然后真的不需要init函数，你可以将这个逻辑添加到构造函数本身：

var MyClass = function(){
    this.customer = null;

    this.test = function(data){
      alert('testing');
    };
};
var testClass = new MyClass();
testClass.customer = 'John B';
testClass.test(); //alerts 'testing'

您还可以将方法添加到MyClass.prototype，而不是在构造函数中声明它们。有关两者之间的差异，请参阅Use of 'prototype' vs. 'this' in JavaScript?。

最后，如果你想坚持你的对象文字，你必须使用Object.create：

var myclass = {
    init:function () {
        this.customer = null;
    },
    test : function(data){
      alert('testing');
    }
};

var testClass = Object.create(myclass);
testClass.customer = 'John B';
testClass.test(); //alerts 'testing'

Answer 2

另一个实现，有一些解释：

var MyClass = function() {
    this.customer = null;
};

// Any functions need to be added to the prototype,
// and should use the keyword this to access member fields.
// Doing this allows for a performance gain over recreating a new function definition
// every time we create the object, as would be the case with this.test = function() { ... }
MyClass.prototype.test = function(data){
    alert('testing');
};

// At this point, MyClass is a constructor function with all of it's
// prototype methods set, ready to be instantiated.

var testClass = new MyClass();
testClass.customer = 'John B'; // May also want to consider moving this into the constructor function as a parameter.
testClass.test();

JSFiddle

Answer 3

您必须将测试方法添加为init的原型。像这样......

var myclass = {
    init:function () {
        this.customer = null;
    },
    test : function(data){
      alert(testing);
    },
};

myclass.init.prototype = myclass;

这样所有对象都将继承自myclass对象。

无法在类实例化后调用方法

3 个答案: