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Question

Help to define generic, which accepts Object with type T and K property key in T and returns typed T .

My attempts to figure out how it can be solved with TypeScript 4:

1. It solves the problem, but returns object . It's not good solution.

type WithProp<TKey extends string> = { [K in TKey]: object };

Playground .

2. The same, but returns any . It is better to write something like T[K] , but TS said that TS2536: Type 'K' cannot be used to index type 'T' .

type WithProp<T, K extends string | number> = T & {[keyToSort in K]: any};

Playground .

---

Usage:

const prop = <K extends string | number>(prop: K) => <T>(model: WithProp<T, K>) => model[prop];
const x = {foo: 'bar', baz: 'qux'};

typeof prop('foo')(x) === typeof x.foo;

And calling the helper with any key (except of 'foo' or 'baz' ) on x must throw an error.

Please, help to write a helper.

Answer 1

If I'm understanding the question, then a conditional which checks that T contains a property K is what you want. Try:

type WithProp<T, K extends string|number> = T extends { [k in K]: any } ? T : never

Then used with your code:

function prop<K extends string | number>(prop: K) {
    return function<T> (model: WithProp<T, K>) { 
        return model[prop];
    }
}

const x = { foo: 'bar', baz: 'qux' };

typeof prop('foo')(x) === typeof x.foo; // Works

prop('blah')(x); // Error, argument is not assignable to never

Not the most descriptive error message, but it gets the job done.

For a better error message, we can re-write things slightly and instead of WithProp we can just put the constraint in the template argument list.

function prop<K extends string | number>(prop: K) {
    return function<T extends { [k in K]: any }> (model: T) { 
        return model[prop];
    }
}

Now we'll get:

prop('blah')(x) // Error, Property 'blah' is missing in type '{ foo: string; baz: string; }' but required in type '{ blah: any; }'