Typescript: Mapped Types: Class to Interface without methods

Question

Is it possible to write a Mapped Type that transforms a Class into an Interface minus the class's methods or any properties assigned to the prototype. EG,

class Dog {
    name: string;
    age: number;
    bark() {}
    get dogYears(): number {
        return this.age * 7;
    }
}

type InterfaceOf<T> = { /* ??? */ };
type IDog = InterfaceOf<Dog>;
// equivalent to
interface IDog {
    name: string;
    age: number;
}

---

Why do I want to do this? I'm looking to "deserialize" json objects into classes. Eg, I run a query to get dogs from the database, afterwards I'd like to instantiate them into class objects, perhaps by using the class-transformer library.

function queryDogs(): Promise<{ name: string, age: string }[]>;
function deserialize<T>(cls: (new() => T), input: InterfaceOf<T>): T;
function deserialize<T>(cls: (new() => T), inputs: InterfaceOf<T>[]): T[];

class Dog {
    @Type(() => String)
    name: string;
    @Type(() => Number)
    age: number;
    bark() {}
    get dogYears(): number {
        return this.age * 7;
    }
}

const dbDogs = await queryDogs();
const dogs: Dog[] = deserialize(Dog, dogs);

It would be nice if the deserialize function knew if the input was the right shape to be deserialized into the Dog class. I was hoping it could look at the Dog class that is given to it to transform it into the appropriate interface.

Answer 1

Interface Based on Class

You can directly generate an interface from a class in TypeScript:

interface DogLike extends Dog {

}

The Angular community is all over this, but beware of some warnings about using classes as interfaces.

The interface that this would generate for you would include properties and methods:

interface DogLike {
    name: string;
    age: number;
    bark(): void;
    dogYears: number;
}

Mapped Type Madness

Now, you can do something really clever/complex/mad with mapped types, based on the mapped types found in this article.

type Remove<T extends string, U extends string> = ({[P in T]: P } & {[P in U]: never } & { [x: string]: never })[T];
type RemoveProperty<T, K extends keyof T> = { [P in Remove<keyof T, K>]: T[P] };

type PartDog = RemoveProperty<DogLike, 'bark' | 'dogYears'>;

This takes the DogLike interface, and removes bark and dogYears.

I included this as you mentioned mapped types. However, it is an insane solution.

Interface

My recommended solution, would be a a simple interface, and perhaps one that isn't named after dogs at all, as the properties are more general:

interface NamedLivingOrganism {
    name: string;
    age: number;
}

Okay, you may not name it exactly like that. However, simple is often best and when you come to change either the Dog class, or the interface that is loosely based on it at some point in the future, the simple interface will prove to be the best way to represent what you need.

Answer 2

It's possible to define the IDog type without methods, as described in TS documentation. But unfortunately it still contains dogYears getter. As I know, it cannot be solved because when dealing with types, there is no difference between fields and getters.

type NonFunctionPropertyNames<T> = {
    [K in keyof T]: T[K] extends Function ? never : K
}[keyof T];
type NonFunctionProperties<T> = Pick<T, NonFunctionPropertyNames<T>>;

type IDog = NonFunctionProperties<Dog>;