Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
5.1k views
in Technique[技术] by (71.8m points)

Initialize empty object in typescript with Record type

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


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...