Infer Typescript class to return the generic type based of generic value

Question

It is difficult to articulate exactly what I want to achieve, so I will post the code. I have a class that should fetch config that should dynamically return the correct config type based on the id specified in the contractor. Question: How should I structure my code solve my problem spesified below?

export type ConfigA = {
  id: "configA";
  name: string;
  propA: string;
};

export type ConfigB = {
  id: "configB";
  name: string;
  propB: string;
};

export type Configs = ConfigA | ConfigB;

export class ConfigService<C extends Configs> {
  private readonly configCollection: CollectionReference<C>;

  constructor(private configName: C["id"]) {
    this.configCollection = db.collection("config") as CollectionReference<C>;
  }

  public async get() {
    const configSnap = await this.configCollection.doc(this.configName).get();
    return configSnap.data();
  }
}

Class callee

const configA = await new ConfigService("configA").get();
const propA = configA.propA;

The Problem:

Property 'propA' does not exist on type 'Configs'.
Property 'propA' does not exist on type 'ConfigB'

Answer 1

You could use a user-defined type guard (aka type predicate) to distinguish between ConfigA and ConfigB in your get() .

Something like this one:

function isConfigA(config: ConfigA | ConfigB): config is ConfigA {
  return (config as ConfigA).propA !== undefined;
}

Answer 2

You could use

new ConfigService<ConfigA>("configA").get()

The issue I think is that the specific subtype cannot be enforced to be inferred from that string parameter alone because ConfigA | ConfigB ConfigA | ConfigB is already a valid generic type. And the rules for type inference are afaik that it will infer the most generic type that fits.

This way you can "help".