how can i extend a typescript class prototype from an array of strings mapping into a single function

Question

i need to extend a prototype, creating methods by mapping a set of strings into a single function. however, i'm having trouble extending a class prototype even for a base case. e.g.:

class Hello {}
Hello.prototype['whatever'] = 1
// [ts] Element implicitly has an 'any' type because type 'Hello' has no index signature.

i read here about index signatures, but not sure how to extend the prototype's definition?

really, i want something pretty simple, similar to as follows:

const methods = ['a', 'b', 'c']
const exampleMethodImplementation = () => 'weeee'
class Hello {
   x: number
}
methods.forEach(name => { Hello.prototype[name] = exampleMethodImplementation })

Jake Holzinger · Accepted Answer · 2018-10-03 20:48:24Z

1

One option is to declare the class with "dynamic" properties.

class Hello {
    [prop: string]: any;
}

Another option is to use Declaration Merging.

type FuncA = () => string;
interface A {
    a1: FuncA;
    a2: FuncA;
    a3: FuncA;
}

type FuncB = (value: string) => string;
interface B {
    b1: FuncB;
    b2: FuncB;
    b3: FuncB;
}

class Hello {
    constructor(public x: number) { }
}

interface Hello extends A, B { }

['a1', 'a2', 'a3'].forEach((method: string) => {
    Hello.prototype[method as keyof A] = function (): string {
        return 'FuncA';
    };
});

['b1', 'b2', 'b3'].forEach((method: string) => {
    Hello.prototype[method as keyof B] = function (value: string): string {
        return `FuncB: ${value}`;
    };
});

const hello = new Hello(1);
console.log(hello.a1());        // FuncA
console.log(hello.b1('foo'));   // FuncB: foo

But you'll need to make sure that the interface properties and array elements stay synchronized.

edited Oct 3, 2018 at 20:48

answered Oct 2, 2018 at 23:32

Jake Holzinger

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2 Comments

cdaringe Over a year ago

thanks. though this could work, practically it does not if you use other properties on your class that don't fit that fn signature. i will update the example class with another property to demonstrate.

Jake Holzinger Over a year ago

Correct, but you could use any instead of the function signature. I updated my answer with a "more complete" solution, but it does require a bit of repetition.

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