I'm attempting to create a custom type that can remove specific properties from an object, unlike the Pick utility.
The desired functionality is as follows:
type ObjectType = {
key1: number,
key2: string,
key3: boolean,
key4: number[]
}
let obj: Remove<ObjectType, 'key2' | 'key4'>
In this scenario, the type of obj should be: { key1: number, key3: boolean }
In TypeScript version 3.5 and above, you can utilize the Omit helper to achieve the functionality you have described.
type ObjectType = {
key1: number,
key2: string,
key3: boolean,
key4: number[]
}
let obj: Omit<ObjectType, 'key2' | 'key4'> // results in type { key1: number, key3: boolean }
If you are using an older version of TypeScript (>2.8), Eduard suggests a combination of Pick and Exclude.
To simplify this process, I typically create a helper type to shorten the code (this definition is directly from TS 3.5):
type Omit<T, K extends keyof any> = Pick<T, Exclude<keyof T, K>>;
To remove the properties key2 and key4, you can use the following method:
type NewType<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>
let myObj: NewType<MyObjectType, 'key2' | 'key4'>
Explanation:
The code Exclude<keyof T, K> will result in 'key1' | 'key3'
After applying
Pick<T, Exclude<keyof T, K>>{ key1: number, key3: boolean }
One way to exclude a specific property from an interface is by using the 'Omit' utility type.
import { Omit } from "utility-types"
interface IHouse {
rooms: number
gardenSize: number
toilets: number
}
type Appartment = Omit<IHouse, "gardenSize">
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