How can I incorporate a new function into a TypeScript class that already exists?

In my current angular 2 cli project, I am faced with the task of defining a plugin that does not have its own type definition. This plugin relies on a main library that already has its own typed definitions and is functioning properly.

I have two files; the first one contains:

LIBRARY TYPES FILE A

export class A extends B {
    constructor(...);
    methodX(): void;
}

Now, I need to add a new method for my plugin so that my class looks like this:

export class A extends B {
        constructor(...);
        methodX(): void;
        methodY(): void;
    }

The challenge lies in adding this new method in a separate file without creating a new class altogether.

If I try putting it in either of these files:

PLUGIN TYPES FILE B

export class A extends B {
    constructor(...);
    methodX(): void;
}

or

PLUGIN TYPES FILE B

export class A extends B {
        constructor(...);
        methodX(): void;
        methodY(): void;
}

It doesn't seem to work. Can anyone provide guidance on how I can successfully overwrite or extend a class with a new method?

Thank you.

Answer №1

If you're facing the challenge of adding a method to a class from a different module in TypeScript, the "Declaration Merging > Module Augmentation" section in the TypeScript documentation could have the answer:

https://www.typescriptlang.org/docs/handbook/declaration-merging.html#module-augmentation

For your specific scenario where class A is exported from file1.ts and you wish to include methodY() from another module file2.ts, you can try the following approach:

//// 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() {}

Answer №2

You have the option to create an interface with a new method and update the prototype for implementation.

Here is an example:

class B { }

class A extends B {
    constructor() {
        super();
    }
    methodX(): void { };
    methodY(): void { };
}


interface B {
    newMethod(): void;
}

B.prototype.newMethod = function () { console.log('a') };

This way, you can maintain proper typing during execution.

new A().newMethod();

I have provided a demonstration in the TypeScript playground here.

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