TypeScript: remove key from type/subtraction type

Update for TypeScript 3.5: The Omit<Type, Keys> utility type is now available. Please see Mathias' answer for an example usage.

Old Answer: Since TypeScript 2.8 and the introduction of Exclude, It's now possible to write this as follows:

type Without<T, K> = {    [L in Exclude<keyof T, K>]: T[L]};

Or alternatively, and more concisely, as:

type Without<T, K> = Pick<T, Exclude<keyof T, K>>;

For your usage, you could now write the following:

type ExcludeCart<T> = Without<T, "cart">;

While this has been correctly answered, I wanted to point out that TypeScript 3.5 did add an Omit<T, E> type.

type NoCart = Omit<{foo: string, bar: string, cart: number}, "cart">;

This results in the {foo: string, bar: string} type.

While there isn't a built-in subtraction type, you can currently hack it in:

type Sub0<    O extends string,    D extends string,> = {[K in O]: (Record<D, never> & Record<string, K>)[K]}type Sub<    O extends string,    D extends string,    // issue 16018    Foo extends Sub0<O, D> = Sub0<O, D>> = Foo[O]type Omit<    O,    D extends string,    // issue 16018    Foo extends Sub0<keyof O, D> = Sub0<keyof O, D>> = Pick<O, Foo[keyof O]>

In the question's case, you would do:

type ExcludeCart<T> = Omit<T, 'cart'>

With TypeScript >= 2.6, you can simplify it to:

/** * for literal unions * @example Sub<'Y' | 'X', 'X'> // === 'Y' */export type Sub<    O extends string,    D extends string    > = {[K in O]: (Record<D, never> & Record<string, K>)[K]}[O]/** * Remove the keys represented by the string union type D from the object type O. * * @example Omit<{a: number, b: string}, 'a'> // === {b: string} * @example Omit<{a: number, b: string}, keyof {a: number}> // === {b: string} */export type Omit<O, D extends string> = Pick<O, Sub<keyof O, D>>

test it on the playground