用 Record 类型初始化 typescript 中的空 object

Question

How to define and initialize an object that can be empty.

With types

type Plan = 'plan1' | 'plan1';

interface IPlan {
    name: string
}

When I tried to initialize an empty object, I'm getting an error

const plans: Record<Plan, Readonly<IPlan> = {}; // **ERROR HERE**

plans.plan1 = {
    name: 'Plan #1'
}

Property 'plan1' is missing in type '{}' but required in type 'Record<"plan1", Readonly>'.

Playgro

Answer 1

You could do something like:

type Plan = 'plan1' | 'plan2';

interface IPlan {
    name: string
}

type PlansRecord = Record<Plan, Readonly<IPlan>>
const plansRecord = {} as PlansRecord

console.log({plansRecord})

Output: [LOG]: { "plansRecord": {} }

Demo

Answer 2

Simply use the Partial utility type: Partial<Type>

type Plan = 'plan1' | 'plan1';

interface IPlan {
    name: string
}


const plans: Partial<Record<Plan, IPlan>> = {}; // no error

plans.plan1 = {
    name: 'Plan #1'
}

The downside of this approach is that now all the properties of your interface are optional. But since you want it to instantiate without the required property, that is the only way.

Playground Link

Another idea might be using the Omit utility type: Omit<Type, Keys>

interface Plan {
  name: string;
}

type IPlan = Omit<Plan , "name">;

const plans: IPlan = {};

So, again, you can instantiate without the required properties.

Playground Link

Answer 3

Looking atthe documentation for Typescript's Record , it looks like they require a more elaborate object definition:

const plans: Record<Plan, Readonly<IPlan>> = {plan1: {name: 'foo'}, plan2: {name:'bar'}}

or:

  const plans: Record<Plan, Readonly<IPlan>> = {plan1: {name: 'foo'}}

If you're just trying to create an element with types, you could just use a class or interface with Optional Properties (see here) . for example:

interface IPlan {
    name?: string;
}