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How do I write out many variables in a type in short hand, from a JSON response, in typescript

I have a relatively trivial question, say I need to create a new type from a JSON Response

My JSON response (lets call it Label Response) is as follows

{
id1: "someString"
someField: "someString"
someMoreFields: "someString"
EvenMorefields: "someString"
EvenMoreMoreFields: "someString"
SoManyManyFields: "someString"
EndlessFields: "someString"
}

Now if I choose to create a type from this response it would be as follows (correct me if i am wrong)

export type LabelResponse =
      {
       Fields: string;
       FieldsMore:string;
       MoreMoreFields:string;
       MoreMoreMoreFields:string;
      }

This is very verbose Since the more IDs, I have the more fields I will need and all the Ids are of the same type.

Is there a way to possible write this in short hand, for example, in my function parameter, instead of declaring a type for it

For example instead of

function myFunction(label:LabelResponse)

Would it be possible to declare my type in the parameter brackets ? Like this:

function myFunction(label:DeclareMyTypeHereSomehow?)

about 4 years ago · Juan Pablo Isaza

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Answer 1 · 2022-07-13T10:39:35.083Z

You can create a Record type include a string key and a value key like so:

type LabelResponse = Record<string, string>

const data: LabelResponse = {
    id1: "someString",
    id2: "someString",
    someRandomString: "someString",
    someWeirdString: "someString",
}

Playground

Answer 2 · 2022-07-13T10:39:35.093Z

You can shorten the type definition using template literal types like following:

export type LabelResponse = { [K in `id${1|2|3|4}`]: string; }

This will give a type as:

type LabelResponse = {
    id1: string;
    id2: string;
    id3: string;
    id4: string;
}

Playground.

Or if you don't care about the actual range, you can use id${number}:

export type LabelResponse = { [K in `id${number}`]: string; }

// errors since key doesn't start with id
const x:  LabelResponse = {
  'x1': 'hello'
}

// works
const y: LabelResponse = {
  'id2': 'world'
}

// not works since it expects a number after id
const z: LabelResponse = {
  'idx': '!'
}

Playground

If you don't care about the key at all, then you can just do:

type LabelResponse = {
  [K: string]: string
}

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How do I write out many variables in a type in short hand, from a JSON response, in typescript

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Andres GPT