I have the following type defined:
export type colors = "red" | "blue" | "green"
export interface details {
name: string;
id: string;
}
Is there a way to define a type object that requires all 3 to be defined, without having to loop through each explicitly?
export interface allColors {
red: details;
blue: details;
... etc.
}
as opposed to this type which makes the keys optional?
export type allColors = {
[key in colors]: details;
};
You can achieve this by using type alias and Record type.
export type colors = "red" | "blue" | "green"
export interface details {
name: string;
id: string;
}
export type allColors = Record<colors, details>
// Usage
const x: allColors = { // type error, missing properties: 'blue', 'green'
red: {
name: 'red',
id: '1'
}
}
Suppose you want to override the type of certain keys, say you want green: number instead of green: details, you can achieve this using Exclude and Intersection Types:
export type allColors = Record<Exclude<colors, 'green'>, details> & {green: string}
// Usage
const x: allColors = {
red: {
name: 'red',
id: '1'
},
green: 'hello' // No error
}
interfacewould require all of them to be set."strict": truein tsconfig.json compilerOptions and are running a relatively recent version of TypeScript