Objects with unknown property names

[b-zurg]

Very often I'll encounter situations where I need to type an object that has unknown property names (e.g. an id or a key) but the shape of its value is known.

e.g.

type Point = { x: number; y: number};
interface PointMap {
  [key: string]: Point;
}

I can't really figure out how to accomplish that with computed-types.

[moshest]

Hi, thanks for bring this up!

I still didn't have the time to look into it, I think the best solution will be to implement some Record<K, V> validator to solve it.
If this urgent, you can always create your own validator like this:

const PointSchema = Schema({ x: number, y: number });

const PointMapSchema = object.test((obj) => Object.values(obj).every(PointSchema));

[b-zurg]

Thanks for responding! It's not urgent but good to know about the example you provided! Having some sort of Record<K, V> seems perfect.

[moshest]

Proposed interface:

import { Record, string, number } from 'computed-types';

const PointMap = Record(string, { x: number, y: number });

[b-zurg]

In a Record<K,V> approach I guess the only valid index types for K would be a string or string union. The V however could be any valid schema. The above example is an inline schema definition I'm guessing but it would be nice to be able to also provide schemas defined elsewhere.

[moshest]

The key validator will be any value that can as as an object property (number, string or symbol). You will be able to provide any validator such as: Schema.enum(MyKeys), Schema.either('foo', 'bar') etc..

[bradenmacdonald]

Hi, here is a generic version that I created today, which seems to work well and will also create the correct Record<K, V> TypeScript type when used with Type<>.

type Validator<T> = (value: any) => T;  // I can't find if/where this type is exported from computed_types?

export const Record = <KeyType extends string | number | symbol, ValueType>(keySchema: Validator<KeyType>, valueSchema: Validator<ValueType>) => {
    return (object
        .test((obj) => Object.values(obj).every(valueSchema) && Object.keys(obj).every(keySchema))
        .transform(obj => obj as Record<KeyType, ValueType>)
    );
}

You will be able to provide any validator such as: Schema.enum(MyKeys), Schema.either('foo', 'bar') etc..

Yep, that is working.

[marcelbeumer]

Are there plans to release this?

[moshest]

I don't have the time right now. You're welcome to create a PR.

[moshest]

Released in v1.11.0

[marcelbeumer]

Awesome, thank you!