Question

创建对象时遇到问题。控制台说最后一行有问题。请告诉它应该如何，我更熟悉Java，所以这对我来说有点混乱。

var dog = {
        name:"Dog",
        age:"11",
        getName : function() {
            alert(this.name);       
        }
    }   

    function Dog(name, age) {
        this.name = name;
        this.age = age; 
    }

    var d1 = new Dog("Rex", 8);
    d1.getName();

Answer 1

您的dog只是一个简单的对象字面值，
这意味着您的getName与其绑定，而不是Dog 类。

您可以将该功能改为method Dog：

＆＃13;

/*var dog = {
  name:"Dog",
  age:"11",
  getName : function() {
    alert(this.name);       
  }
}*/ 

function Dog(name, age) {
  this.name = name;
  this.age = age; 
}

Dog.prototype.getName = function() {
  console.log( this.name );
}

var d1 = new Dog("Rex", 8);
d1.getName(); // "Rex"

＆＃13;

这是一个使用settings＆＃34;默认值＆＃34;

的变体

＆＃13;

function Dog() {
  this.name = "Dog"; // Default name
  this.age = 11;     // Default age
}

Dog.prototype.getName = function() {
  console.log( this.name );
}

var d1 = new Dog();
d1.name = "Rex";    // Override default name
d1.getName(); // "Rex"

＆＃13;

Answer 2

您可以使用带语法糖的类在ES6中正确创建对象在你的例子中，这样写：

'use strict';
class Dog{
  constructor(name, age){
    this.name = name;
    this.age = age;
  }
  getName(){
    console.log(this.name);
  }
}

let doggy = new Dog("krypto", 125);
doggy.getName();

Answer 3

JavaScript中的传统OO

function Dog(name, age) {
    this.name = name || "Dog";// if the name is not given, it defaults to "Dog"
    this.age = age || "11"; 
}

Dog.prototype.getName =  function() {
    alert(this.name); 
}

var d1 = new Dog("Rex", 8);
d1.getName();

JavaScript中的更多显式OO

function createDog(name, age) {
    // create a new dog and return it
    var dog = {
        name: name || "Dog",// if the name is not given, it defaults to "Dog"
        age: age || "11"
    };
    return dog;
}

createDog.getName =  function(dog) {
    // explicit dog as 1st parameter
    alert(dog.name); 
}

//createDog is a normal function that returns something, no "new" needed
var d1 = createDog("Rex", 8);
createDog.getName(d1);

JavaScript OOP：Uncaught TypeError：不是函数

3 个答案: