Exclude property from type

I'd like to exclude a single property from the type. How can I do that?

For example I have

interface XYZ {
  x: number;
  y: number;
  z: number;
}

And I want to exclude property z to get

type XY = { x: number, y: number };

For versions of TypeScript at or above 3.5

In TypeScript 3.5, the Omit type was added to the standard library. See examples below for how to use it.

For versions of TypeScript below 3.5

In TypeScript 2.8, the Exclude type was added to the standard library, which allows an omission type to be written simply as:

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

For versions of TypeScript below 2.8

You cannot use the Exclude type in versions below 2.8, but you can create a replacement for it in order to use the same sort of definition as above. However, this replacement will only work for string types, so it is not as powerful as Exclude.

// Functionally the same as Exclude, but for strings only.
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>>

And an example of that type in use:

interface Test {
    a: string;
    b: number;
    c: boolean;
}

// Omit a single property:
type OmitA = Omit<Test, "a">; // Equivalent to: {b: number, c: boolean}

// Or, to omit multiple properties:
type OmitAB = Omit<Test, "a"|"b">; // Equivalent to: {c: boolean}

Omit

single property

type T1 = Omit<XYZ, "z"> // { x: number; y: number; }

multiple properties

type T2 = Omit<XYZ, "y" | "z"> // { x: number; } 

properties conditionally

type Keys_StringExcluded<T> = 
  { [K in keyof T]: T[K] extends string ? never : K }[keyof T]

type XYZ = { x: number; y: string; z: number; }
type T3a = Pick<XYZ, Keys_StringExcluded<XYZ>> // { x: number; z: number; }

type T3b = { [K in keyof XYZ as XYZ[K] extends string ? never : K]: XYZ[K] } 
// { x: number; z: number; }

properties by string pattern

type OmitGet<T> = {[K in keyof T as K extends `get${infer _}` ? never : K]: T[K]}

type XYZ2 = { getA: number; b: string; getC: boolean; }
type T4 = OmitGet<XYZ2> //  { b: string; }

Note: Above template literal types are supported with TS 4.1.
Note 2: You can also write get${string} instead of get${infer _} here.

Pick, Omit and other utility types

How to Pick and rename certain keys using Typescript? (rename instead of exclude)

Playground

With typescript 2.8, you can use the new built-in Exclude type. The 2.8 release notes actually mention this in the section "Predefined conditional types":

Note: The Exclude type is a proper implementation of the Diff type suggested here. [...] We did not include the Omit type because it is trivially written as Pick<T, Exclude<keyof T, K>>.

Applying this to your example, type XY could be defined as:

type XY = Pick<XYZ, Exclude<keyof XYZ, "z">>

I've found solution with declaring some variables and using spread operator to infer type:

interface XYZ {
  x: number;
  y: number;
  z: number;
}

declare var { z, ...xy }: XYZ;

type XY = typeof xy; // { x: number; y: number; }

It works, but I would be glad to see a better solution.

In Typescript 3.5+:

interface TypographyProps {
    variant: string
    fontSize: number
}

type TypographyPropsMinusVariant = Omit<TypographyProps, "variant">