Typescript 2 Generic types dependent on each other

Question

I have two types of config one for auth and one for general config which doesn't have any secret variables I used to have two functions but now I'm trying to have only one function but I wasn't able to implement this code with errors, I couldn't find what I'm doing wrong

interface config {
    config: string;
    config_example: string;
} 

interface auth {
    auth: string;
    auth_example: string;
}

type configType = "auth" | "config";
type getType<C extends configType> = C extends "auth" ? keyof auth : C extends "config" ? keyof config : never;

const auth: auth = {
    auth: '',
    auth_example: ''
}

const config: config = {
    config: '',
    config_example: ''
}

function getConfig<
    C extends configType,
    K extends keyof getType<C>
>(config: C, key: K): getType<C>[K] {
    switch (config) {
        case "auth":
            return auth[key];
        case "config":
            return config[key];
        default:
            throw new Error();
    }
}

here's the typescipt playground code of this

If I put //@ts-ignore to the errors, the IDE correctly determines the types and everything but I don't want to implement this with //@ts-ignore I don't even know if this is possible to do that but any ideas?

Lesiak · Accepted Answer · 2023-02-10 18:58:47Z

1

See Wishlist: support for correlated union types #30581

const allConfigs = {
  "auth": auth,
  "config": config
}

function getConfig<
    CONFIG_NAME extends keyof typeof allConfigs,
    CONFIG_KEY extends keyof typeof allConfigs[CONFIG_NAME]
>(config: CONFIG_NAME, key: CONFIG_KEY): typeof allConfigs[CONFIG_NAME][CONFIG_KEY] {
    return allConfigs[config][key];
}

Code in TS Playground

edited Feb 10, 2023 at 18:58

answered Feb 9, 2023 at 19:28

Lesiak

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4 Comments

jcalz Over a year ago

Uh... why the switch/case statement? Does this version not work for some reason?

Lesiak Over a year ago

@jcalz Indeed - even in the unlikely case when someone passes unknown config name as any, TypeError "Cannot read properties of undefined" is thrown - no need to check it ourselves. I updated the answer.

jcalz Over a year ago

... did you? I don't see it.

Lesiak Over a year ago

@jcalz Now I'm pretty sure the change is applied.

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