Question

"use strict";
var result = document.getElementById('result');

function Person() {
  //STATIC CALL
  Person.callfunc();

}

//STATIC PROPERTY
Person.age = 20;
//STATIC METHOD
Person.callfunc = function() {
  result.innerHTML += Person.age + "<br>";
};

var obj = new Person();
result.innerHTML += obj.age + "<br>";

<div id='result'></div>

obj无法访问年龄。但是作为父母的孩子应该。我来自Java，刚接触JS。

Answer 1

使用Object.prototype怎么样？应该会更好。

"use strict";
var result = document.getElementById('result');

function Person() { }

//STATIC PROPERTY
Person.prototype.age = 20;

var obj = new Person();
result.innerHTML += obj.age + "<br>";

<div id='result'></div>

Answer 2

Person.age和Person.prototype.age之间的区别：

通过Person.age age是函数对象Person()的属性
by Person.prototype.age age是类对象Person()的属性

请阅读文档： Object.prototype：

通过原型链接，所有对象可以看到
更改（所有对象都可以看到对象原型对象的更改），除非沿着原型链进一步覆盖发生这些更改的属性和方法。这提供了一种非常强大的机制，尽管具有潜在的危险性，它可以覆盖或扩展对象的行为。

function Person()
{
    //static call from function object property
    Person.callfunc1();
    
    this.weight = 71; //class object property
    
    this.callfunc2 = function()
    {
        console.log('weight: ', this.weight); // 'weight: 71' because weight is in class object and 'this' is here class object
    };
    
    //call from class object property
    this.callfunc2();
}

//static property from function object
Person.age = 25;
Person.prototype.name = 'Peter';

//static method from function object
Person.callfunc1 = function()
{
    console.log(Person.age); // 25
    console.log('weight: ', this.weight); // 'weight: undefined' because weight is in class object, but 'this' is here function object
};

var obj = new Person(); // obj is not a child of Person, this in the instance of it
console.log('obj.age ', obj.age); // 'obj.age undefined' because it is NOT in class object
console.log(obj.name); // 'Peter' because it is in class object
console.log('obj.callfunc1 ', obj.callfunc1); // 'obj.callfunc1 undefined' because it is NOT in class object
console.log(obj.weight); // 71 because it is in class object

Answer 3

简单地说，实例（在这种情况下为obj）无法通过原型链访问构造函数（在这种情况下为Person）的静态属性和方法。但是，您可以通过.constructor方法访问它们。因此，以下代码将起作用：

"use strict";
var result = document.getElementById('result');

function Person() {
  //STATIC CALL
  Person.callfunc();

}

//STATIC PROPERTY
Person.age = 20;
//STATIC METHOD
Person.callfunc = function() {
  result.innerHTML += Person.age + "<br>";
};

var obj = new Person();
result.innerHTML += obj.constructor.age + "<br>";

<div id='result'></div>

但是我不知道你为什么要这样做。

为什么子代无法在js中访问父代的静态变量/方法？

3 个答案: