typescript 向打字稿中的现有类添加方法？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/42999765/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Add a method to an existing class in typescript?
提问by ackuser
I am working in an angular 2 cli project in which I have to create a definition of a plugin, because it doesn't exists its typed. This plugin depends of a main library that has already its own typed and it works.
我正在一个 angular 2 cli 项目中工作，我必须在其中创建插件的定义，因为它不存在其类型。这个插件依赖于一个已经有自己的类型并且可以工作的主库。
Anyway, I have two files the main one with
无论如何，我有两个文件，主要的一个
LIBRARY TYPES FILE A
库类型文件 A
export class A extends B {
constructor(...);
methodX(): void;
}
And I would need to add a new method for my plugin so my class would be like
我需要为我的插件添加一个新方法，这样我的班级就会像
export class A extends B {
constructor(...);
methodX(): void;
methodY(): void;
}
The point is that I need to add it in a separate file. The problem is adding a method to an existent class without creating a new one
关键是我需要将它添加到一个单独的文件中。问题是向现有类添加方法而不创建新方法
If I put
如果我把
PLUGIN TYPES FILE B
插件类型文件 B
export class A extends B {
constructor(...);
methodX(): void;
}
or
或者
PLUGIN TYPES FILE B
插件类型文件 B
export class A extends B {
constructor(...);
methodX(): void;
methodY(): void;
}
It doesn't work, does anyone how can I achieve overwriting a class or extending it with a new method that?
它不起作用，有没有人如何实现覆盖类或使用新方法扩展它？
Thanks
谢谢
回答by toskv
You could do it by making an interface with the new method and modifying the prototype.
您可以通过使用新方法创建接口并修改原型来实现。
Something like this:
像这样的东西：
class B { }
class A extends B {
constructor() {
super();
}
methodX(): void { };
methodY(): void { };
}
interface B {
newMethod(): void;
}
B.prototype.newMethod = function () { console.log('a') };
This allows you do have proper typing when doing.
这允许你在做的时候有正确的打字。
new A().newMethod();
I made a playground example here.
我在这里做了一个游乐场的例子。
回答by Bünyamin Sar?gül
You can do it directly as A.prototype.functionName = function(){...}
你可以直接做 A.prototype.functionName = function(){...}
Here is a plunker: http://plnkr.co/edit/6KrhTCLTHw9wjMTSI7NH?p=preview
这是一个plunker：http://plnkr.co/edit/6KrhTCLTHw9wjMTSI7NH?p=preview
回答by Serban Stokker
The "Declaration Merging > Module Augmentation" section from the TypeScript docs seems to offer the solution:
TypeScript 文档中的“声明合并 > 模块增强”部分似乎提供了解决方案：
https://www.typescriptlang.org/docs/handbook/declaration-merging.html#module-augmentation
https://www.typescriptlang.org/docs/handbook/declaration-merging.html#module-augmentation
In your case, if class Ais exported from file1.ts, and you want to add methodY()to that class within a different module file2.ts, then try this:
在您的情况下，如果class A从 导出file1.ts，并且您想methodY()在不同的模块中添加到该类file2.ts，请尝试以下操作：
//// file1.ts
export class A extends B {
constructor(...);
methodX(): void;
}
//// file2.ts
import { A } from "./file1";
declare module "./file1" {
interface A {
methodY(): void;
}
}
A.prototype.methodY = function() {}
