Overriding Interface Property Type Defined in Typescript D.Ts File

Overriding interface property type defined in Typescript d.ts file

I use a method that first filters the fields and then combines them.

reference Exclude property from type

interface A {
    x: string
}

export type B = Omit<A, 'x'> & { x: number };

for interface:

interface A {
    x: string
}

interface B extends Omit<A, 'x'> {
  x: number
}

How to override type properties in TypeScript

Create a helper type:

type Overwrite<T, U> = Pick<T, Exclude<keyof T, keyof U>> & U;

Usage:

type HexColorLine = Overwrite<Line, { color: number }>

Overriding interface property type from exported type

If I understood your question correctly, I think this is the type you want:

You don't need to supply generic type parameters when you are choosing the same ones as the defaults.

TS Playground

interface Place extends Omit<Collections.Places.Entity, 'assignedPlaces'> {
  assignedPlaces?: NewPlaces[];
  bounds: Bounds;
  center: Position;
  id: string;
}

declare const place: Place;
place.assignedPlaces; // NewPlaces[] | undefined

Override the properties of an interface in TypeScript

UPDATE, 2018-08

TypeScript 2.8 introduced Exclude<T, U> which behaves like the Diff<T, U> defined below for all types (not just key types). You should definitely use Exclude<> instead of Diff<> if you are using TypeScript 2.8 or above.

Also, in TypeScript 2.9, keyof any was expanded from string to string | number | symbol, so the below Diff<T, U> caused errors which can be fixed by changing Diff<T extends string, U extends string> to Diff<T extends keyof any, U extends keyof any>. This change has been made below.

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ORIGINAL ANSWER, 2018-03

Yes, conditional types will enable this, although you can get this behavior without them as well.

The idea is to pull properties out of the original interface and then replace them with new ones. Like this:

type Diff<T extends keyof any, U extends keyof any> = 
  ({ [P in T]: P } & { [P in U]: never } & { [x: string]: never })[T];
type Overwrite<T, U> = Pick<T, Diff<keyof T, keyof U>> & U;

interface OriginalInterface {
  title?: string;
  text?: string;
  anotherProperty?: SomeType;
  // lots of other properties
}
interface Extension {
  title?: string | ReactElement<any>;
  text?: string | ReactElement<any>;
}

interface ExtendedInterface extends Overwrite<OriginalInterface, Extension> {};

const ext: ExtendedInterface = {};  // okay, no required properties
ext.text; // string | ReactElement<any> | undefined
ext.title; // string | ReactElement<any> | undefined
ext.anotherProperty; // SomeType | undefined

EDIT: I changed the definition of Overwrite<T,U> to respect the optional/required status of properties from T whose keys are not present in U.

This has the behavior you want, I think. Hope that helps; good luck!

Access dynamic property of an interface in Typescript

type X = Person[keyof Person]; 

Will return type:

type X = {
  name: string;
  alias: string;
  active: string[];
  values: string[];
}

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