Typescript select string properties

Question

I'm trying to figure out how to correctly type a show function that would take an object T and a key K for which T[K] can be guaranteed to have a toString() method implemented.

Here is my attempt using mapped types

type ToStringablePropertyKeys<T> = keyof {
    [K in keyof T]: { toString(): string }
}

function show<T, K extends ToStringablePropertyKeys<T>>(t: T, k: K): string {
    return t[k].toString()
}

But the compiler says Property 'toString' does not exist on type 'T[K]'.

What am I missing here? How do I convince tsc that toString is in fact there by definition of K?

How is that useful? Every primitive value and object has a `toString` method in JavaScript. — Clashsoft, Jun 22 '21 at 14:18
Since the compiler complains about `toString` not being there, I'm assuming there are valid properties that don't have `toString`. Either way, I might want to apply the same pattern to something less common than `toString`: what if I want to guarantee that a `T[K]` is of a given type `V`? — francoisr, Jun 22 '21 at 14:42
@Clashsoft `null` doesn't have a `toString()`... nor `undefined`. — Andrei Tătar, Jun 22 '21 at 14:58
Please see [the answer](https://stackoverflow.com/a/61765012/2887218) to the question this duplicates for more info; translating that code here yields [this](https://tsplay.dev/wQAojw)... good luck! — jcalz, Jun 22 '21 at 14:59

score 6 · Answer 1 · answered Jun 22 '21 at 14:52

6

An alternative approach:

function show<K extends string, T extends {[key in K]:{toString(): string}}>
(t: T, k: K ): string {
return t[k].toString()
}

answered Jun 22 '21 at 14:52

Nadia Chibrikova

4,916
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2

nice thinking outside the box! – Andrei Tătar Jun 22 '21 at 14:54
this is f... nice solution! upvote! – spirytus Jun 19 '22 at 11:02

score 2 · Answer 2 · answered Jun 22 '21 at 14:51

Here's an almost ideal solution. Though typescript can enforce the keys to be only the ones that expand on {toString():string}, it's not smart enough to infer that from there on, all T[k] will have toString():string...

type ToStringable = { toString(): string };

type ToStringablePropertyKeys<T> = {
    [k in keyof T]: T[k] extends ToStringable ? k : never
}[keyof T];

function show<T, K extends ToStringablePropertyKeys<T>>(t: T, k: K): string {
    return (t[k] as any).toString(); // it seems typescript is not smart enough to infer that all t[k] will have toString
}

const testObject = {
    string: 'test',
    number: 123,
    null: null,
    undefined: undefined,
};

show(testObject, 'number');
show(testObject, 'string');
show(testObject, 'invalid-prop'); //this fails
show(testObject, 'null'); //this fails
show(testObject, 'undefined'); //this fails

https://www.typescriptlang.org/play?ts=4.3.4#code/C4TwDgpgBAKg9gZWAJwJYDsDmBDARgG2gF4oBvKYRFDTACgEoAuKAZ2qygF8BuAKF9CRYVNFjyEACsjiRkoANIQQLADwwAfFBKleUPVADaAaygYoRpXABmsALrMYx21AgAPYBHQATFsKSicAmgAfnMoZnQIADcIZF5OY0sbGFs+XisAV3QAY2BUOHRWAAs4AHc1ABooeRd3Tx8-dkDJaVkFJVUNdVpgByqjZnkmViayXX1kCGAM5EKepyhsX2x0EHoAOkp-GgZuKAB6fdNgVggIAFtfQQgWbLQwE9RfdDgTlnPsORcXjMwiijgpnQVliFCK2BO2Hw+AoC1KqGhUHBMQB2yw8X42QKbAoN2AAHlcAArCC5LRjfQjALMADkHjYNIq4z06Ay51wsWYAEYAEwAZiZlNZ0IiGWhgv0WS8ECsGAgXmYUplcq8TJ4-BYJVKPTxhJJuSqNNZ7NiNPofE1ZR1bD1pOAhrYATNFq11oJxLthowUShqC8AFowK1nQd9sAik8oFZsAiWLxLdr6e79faoEaxfgQ4dw5Ho7H466k7aDWmlbLIl4s2GI748-g40A

Ok, so it seems a cast is necessary after all, I thought I was missing something. — francoisr, Jun 22 '21 at 14:58

Typescript select string properties

2 Answers2