Javascript Question posted in
A very good W3school tutorial can be found here.

Javascript – How to make a key optional in Union Type

I have a type in my TypeScript project

export type Types = "APPLE" | "MANGO" | "BANANA";

Here I want the BANANA to be optional, i.e.,

If I have created a mapping using the keys in Types for example

const MAPPING: { [key in Types]: string } = {
 APPLE: "Here is Apple",
 MANGO: "Here is Mango",
};

This code gives error it’s asking to define mapping for each key if I don’t do it gives error:

Property 'BANANA' is missing in type '{ APPLE: string; MANGO: string; }' but required in type '{ APPLE: string; MANGO: string; 
 BANANA: string; }'

To avoid these types of errors, I want to make the BANANA optional also I don’t want to change the MAPPING itself, it should remain as [key in Types]. I understand that either using Exclude or Types? would work but this requires changes in a lot of files.

Is there any other way to achieve this?

Tags:

2

Answers


  1. 0 
    const MAPPING: { [key in Types]?: number } = { APPLE: 1 };
    Login or Signup to reply.
  2. 0

    You could use an intersection:

    Playground

    export type Types = "APPLE" | "MANGO" | "BANANA";

type MakeOptional<T extends string, E extends string, V = any> = {
    [K in Exclude<T, E>]: V
} & {
    [K in E]?: V
};

type Mapped = MakeOptional<Types, 'BANANA', string>

const MAPPING: Mapped = {
 APPLE: "Here is Apple",
 MANGO: "Here is Mango",
};
    Login or Signup to reply.
Please signup or login to give your own answer.

Back To Top
Search