Yep, identifying InferType optional attributes

Here's an example of a Yup schema I created to fetch entities known as Parcels:

export const FindParcelsParamsSchema = Yup.object({
  cursor: Yup.number().optional(),
  pageSize: Yup.number().positive().integer().optional(),
});

All fields are optional.

Next, let's look at the interface I'm trying to generate for this schema:

export interface IParcelsFetchingParams extends Yup.InferType<typeof FindParcelsParamsSchema> { }

I believe I should be able to declare this variable like so:

let params: TParcelsFetchingParams = {};

But TypeScript keeps raising this error message:

Type '{ cursor: undefined; pageSize: undefined; }' is missing the following properties from type...

Unless I add all the fields as undefined:

let x: TParcelsFetchingParams = {
  cursor: undefined,
  pageSize: undefined,
}; 

My tsconfigs are set as follows:

"compilerOptions": {
    "target": "es5",
    "module": "commonjs",
    "declaration": true,
    "outDir": "./lib",
    "strict": true
  }

So why exactly am I encountering this issue, and what steps can I take to resolve it?

I expect to be able to declare my variable with either no properties or just one without any problems.

Answer №1

Change the declaration of FindParcelsSchema from const to let.

I found that switching from const to let resolved the issue, although I'm not entirely sure why. You can refer to the documentation here where it mentions using let instead of const for all instances of yup.object.

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