TypeScript: conditional type inside the same type/interface

Question

Im trying to get some conditional typing to work and not sure if what Im thinking is possible to do or not. Here it goes:

type AParams = {
    hello: string

}
type BParams = {
    world: string
}

type Data = {
    name: "a" | "b",
    data: AParams | BParams
}

Depending on name being a or b I want to enforce the data to either have type AParams or BParams, respectively. Any way this can be accomplished with TypeScript?

palaѕн · Accepted Answer · 2020-04-29 10:28:01Z

3

You can use two different interfaces for the two cases like:

interface AData {
    name: "a",
    data: AParams
}

interface BData {
    name: "b",
    data: BParams
}

and then set Data using union types like:

type Data = AData | BData;

answered Apr 29, 2020 at 10:28

palaѕн

74.2k17 gold badges123 silver badges140 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

falinsky · Accepted Answer · 2020-04-29 10:22:38Z

0

Your Data type can be next instead:

type Data = {
    name: "a"
    data: AParams
} | {
    name: "b"
    data: BParams
}

answered Apr 29, 2020 at 10:22

falinsky

7,4454 gold badges34 silver badges59 bronze badges

Collectives™ on Stack Overflow

TypeScript: conditional type inside the same type/interface

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related