Conditional type based on the value of another key

Question

Is there a way to use an interface based on the value of another key?

So, I would like options to use a particular interface if driver equals a particular value:

if driver == 'a' use InterfaceA
if driver == 'b' use InterfaceB

I thought something like this might work but it doesn't.

export interface InterfaceA {}
export interface InterfaceB {}

export interface MyInterface {
  driver: string
  root: string
  options: driver == 'a' ? InterfaceA : InterfaceB
}

Evert · Accepted Answer · 2019-06-03 18:42:13Z

1

There's a few different ways to do this, but I tend to take this approach as it gives me the easiest to read error messages:

export interface InterfaceA {}
export interface InterfaceB {}

type MyInterfaceA = {
  driver: 'a',
  root: string,
  options: InterfaceA,
}
type MyInterfaceB = {
  driver: 'b',
  root: string,
  options: InterfaceB,
}
export type MyInterface = MyInterfaceA | MyInterfaceB;

As far as I know these have to be a type not interface, as interfaces don't support union types. (thank you @jcalz).

It's possible to write the above without duplicating the root property, which can be useful if there are more common properties, but I end up duplicating everything anyway as error messages are a lot clearer when avoiding nested unions and intersects.

edited Jun 3, 2019 at 18:42

answered Jun 3, 2019 at 18:34

Evert

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7 Comments

Jordan Running Over a year ago

It may be helpful to point out that this pattern has a name and documentation: Discriminated unions.

Get Off My Lawn Over a year ago

type MyInterfaceA { is missing = btw

Evert Over a year ago

@GetOffMyLawn Thanks, fixed!

Get Off My Lawn Over a year ago

Thanks! Looks like this is working how I would like!

TmTron Over a year ago

@Evert: why do you use type instead of interface for MyInterfaceA/B? Is there any advantage?

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