Define a list of optional keys for Typescript Record

Question

I want to type an object which can only have keys 'a', 'b' or 'c'.

So I can do it as follows:

Interface IList {     a?: string;     b?: string;     c?: string; } 

They are all optional! Now I was wondering if this can be written with Record in just one line

type List = Record<'a' | 'b' | 'c', string>; 

The only issue is that all keys need to be defined. So I ended up with

type List = Partial<Record<'a' | 'b' | 'c', string>>; 

This works, but I can imagine there is a better way to do this without Partial. Is there an other way to make the keys optional inside Record ?

Answer 1

There is no way to specify the optionality of members of Record. They are required by definition

type Record<K extends keyof any, T> = {     [P in K]: T; // Mapped properties are not optional, and it's not a homomorphic mapped type so it can't come from anywhere else. }; 

You can define your own type if this is a common scenario for you:

type PartialRecord<K extends keyof any, T> = {   [P in K]?: T; }; type List =  PartialRecord<'a' | 'b' | 'c', string> 

Or you can define PartialRecord using the predefined mapped types as well:

type PartialRecord<K extends keyof any, T> =  Partial<Record<K, T>> 

Answer 2

You can create the partial version of your List type:

type PartialList = Partial<List>; 

And you could do it all on one line if you don't want the intermediate type:

type PartialList = Partial<Record<'a' | 'b' | 'c', string>>; 

You might decide that, in the end, the most expressive for your future self is:

type List = {     a?: string;     b?: string;     c?: string; }