Is it possible for Typescript to allow extracted interfaces while excluding properties from another interface?

I've been searching for information on the specific features of this. Despite my efforts on Google, I have been unable to find the details. Any help would be greatly appreciated!

interface Numbers {
    number: number;
    number2: number;
    number3: number;
}

const numbers:Numbers - { number3 } = {
    number: 1,
    number2: 2
};

Answer №1

Update

Since the release of Typescript 3.5, the Omit type is now included in the standard types provided by Typescript.

Initial response

Prior to version 2.8, you could achieve this by utilizing the Exclude conditional type along with the Pick mapped type.

type Omit<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>; 

interface Numbers {
    number: number;
    number2: number;
    number3: number;
}

const numbers:Omit<Numbers, 'number3'> = {
    number: 1,
    number2: 2
};

Prior to version 2.8, the definition of Omit would look like this:

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

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