首頁 > web前端 > js教程 > 主體

如何將物件轉換為實例物件並啟用 JavaScript 陣列中的方法使用?

Patricia Arquette
發布: 2024-10-18 12:22:30
原創
361 人瀏覽過

How to Cast Objects to Instance Objects and Enable Method Use in JavaScript Arrays?

Casting Objects to Class Instances in JavaScript

Question:

How can I cast a generic object array (e.g., [{personName: 'John', animals: []}]) received from a server to a typed array of objects (e.g., [Person(personName: 'John', animals: [])]), allowing for method use (e.g., persons[0].animals[2].run())?

Answer:

Creating class instances in JavaScript requires invoking the constructor with the correct arguments (not just properties). We provide two solutions:

Method 1: Static Method for Cloning

Define a static method on the class that takes objects and creates instances from them, for example:

<code class="javascript">Person.fromJSON = function(obj) {
  // Custom code for Person instances
  return new Person();
};</code>
登入後複製

Method 2: Object Cloning

Clone the properties from the JSON-parsed object to a new instance, for example:

<code class="javascript">var personLiteral = JSON.parse("...");
var personInstance = new Person();
for (var prop in personLiteral)
  personInstance[prop] = personLiteral[prop];</code>
登入後複製

Or, using Object.assign:

<code class="javascript">var personInstance = Object.assign(new Person(), personLiteral);</code>
登入後複製

Example for Your Case:

In your specific case, you have a simple object structure with no arguments and only public properties. To convert to an object instance:

<code class="javascript">var persons = JSON.parse(serverResponse);
for (var i=0; i&lt;persons.length; i++) {
  persons[i] = $.extend(new Person, persons[i]);
  for (var j=0; j&lt;persons[i].animals; j++) {
    persons[i].animals[j] = $.extend(new Animal, persons[i].animals[j]);
  }
}</code>
登入後複製

Note: The run method should likely be added to the Animal.prototype object instead of each instance.

以上是如何將物件轉換為實例物件並啟用 JavaScript 陣列中的方法使用?的詳細內容。更多資訊請關注PHP中文網其他相關文章!

來源:php
本網站聲明
本文內容由網友自願投稿,版權歸原作者所有。本站不承擔相應的法律責任。如發現涉嫌抄襲或侵權的內容,請聯絡admin@php.cn
作者最新文章
熱門教學
更多>
最新下載
更多>
網站特效
網站源碼
網站素材
前端模板