如何将方法添加到（JSON）对象的原型？ [英] How to add methods to a (JSON) object's prototype?
问题描述
假设我从服务器收到一些JSON对象，例如Person对象的一些数据：
Let's say I receive some JSON object from my server, e.g. some data for a Person object:
{firstName: "Bjarne", lastName: "Fisk"}
现在，我想要在这些数据之上的一些方法，例如用于计算fullName：
Now, I want some methods on top of those data, e.g. for calculating the fullName:
fullName: function() { return this.firstName + " " + this.lastName; }
这样我就可以
var personData = {firstName: "Bjarne", lastName: "Fisk"};
var person = PROFIT(personData);
person.fullName(); // => "Bjarne Fisk"
我基本上想在这里做的是为对象添加一个方法原型。 fullName（）方法是通用的，因此不应添加到数据对象本身。喜欢..：
What I basically would want to do here, is to add a method to the object's prototype. The fullName() method is general, so should not be added to the data object itself. Like..:
personData.fullName = function() { return this.firstName + " " + this.lastName; }
...会导致大量冗余;并且可以说是污染数据对象。
... would cause a lot of redundancy; and arguably "pollute" the data object.
将这些方法添加到简单数据对象的当前最佳实践方法是什么？
What is the current best-practice way of adding such methods to a simple data object?
编辑：
稍微偏离主题，但如果上述问题可以解决，则可以做到一些不错的伪 - 模式匹配，如下所示：
Slightly off topic, but if the problem above can be solved, it would be possible to do some nice pseudo-pattern matching like this:
if ( p = Person(data) ) {
console.log(p.fullName());
} else if ( d = Dog(data) ) {
console.log("I'm a dog lol. Hear me bark: "+d.bark());
} else {
throw new Exception("Shitty object");
}
人和<如果 data 对象具有正确的属性，code> Dog 将添加方法。如果没有，返回falsy（即数据不匹配/符合）。
Person and Dog will add the methods if the data object has the right attributes. If not, return falsy (ie. data does not match/conform).
奖励问题：是否有人知道一个使用或启用它的库（即使它变得容易）？它已经是一个JavaScript模式？如果是这样，它叫什么;你有一个详细阐述的链接吗？谢谢：）
BONUS QUESTION: Does anyone know of a library that either uses or enables this (ie makes it easy)? Is it already a javascript pattern? If so, what is it called; and do you have a link that elaborates? Thanks :)
推荐答案
假设您的Object来自一些解析服务器输出以生成Object的JSON库，它将不在一般在其原型中有任何特别之处;并且为不同的服务器响应生成的两个对象将不共享原型链（当然除了Object.prototype;）
Assuming your Object comes from some JSON library that parses the server output to generate an Object, it will not in general have anything particular in its prototype ; and two objects generated for different server responses will not share a prototype chain (besides Object.prototype, of course ;) )
如果你控制所有人员的地方是从JSON创建的，你可以反过来做事：创建一个空Person对象（在其原型中使用类似fullName的方法），并使用从JSON生成的对象扩展它（使用$ .extend，_ .extend，或类似的东西）。
If you control all the places where a "Person" is created from JSON, you could do things the other way round : create an "empty" Person object (with a method like fullName in its prototype), and extend it with the object generated from the JSON (using $.extend, _.extend, or something similar).
var p = { first : "John", last : "Doe"};
function Person(data) {
_.extend(this, data);
}
Person.prototype.fullName = function() {
return this.first + " " + this.last;
}
console.debug(new Person(p).fullName());
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