Troubleshooting the issue of conditional extension in Typescript for "Array or Object" not functioning as anticipated

My goal is to create a TypeScript type generic that has the following structure:

type APIDataShape<T extends { id: unknown } | Array<{ id: unknown }>> =
  T extends Array<any>
    ? Array<{
        id: T[number]["id"];
        attributes: Omit<T, "id">;
      }>
    : {
        id: T["id"];
        attributes: Omit<T, "id">;
      };

The concept behind it is simple: I am looking for an object or an array of objects with an 'id' property, and I want to transform it into a different shape.

However, as I write the code above, I encounter an error on the id: T["id"]; line, which says:

Type '"id"' cannot be used to index type 'T'

This error perplexes me because it should be evident that if T isn't an array, then it must be an object with {id: unknown}. So why does TypeScript claim that it cannot be indexed?

I have tried various approaches to resolve this issue without success. If anyone can shed light on why this error is occurring, I would greatly appreciate it.

(P.S. I understand that using two extend clauses could potentially fix the problem, but I am particularly interested in understanding the underlying reason behind it. It seems to me that even with just one extend, it should work as intended. Hence, I seek clarity on why it failed.)

Thank you.

Answer №1

It's important to specify the data shape

type DataFormat<T extends { id: unknown } | Array<{ id: unknown }>> =
  T extends Array<{ id: unknown }>
  ? Array<{
    id: T[number]["id"];
    attributes: Omit<T, "id">;
  }>
  : T extends { id: unknown } ? {
    id: T["id"];
    attributes: Omit<T, "id">;
  } : never;

Answer №2

When working with Indexed Access, it is important to narrow down the type to a specific concrete type or differentiate within a union using techniques like discrimination. One useful method involves utilizing the infer keyword.

type APIDataShape<T extends {id: unknown} | Array<{id: unknown}>> =
  T extends (infer U)[]
    ? {
        id: U extends {id: infer I} ? I : unknown;
        attributes: Omit<U, "id">;
      }[]
    : T extends {id: infer I}
    ? {
        id: I;
        attributes: Omit<T, "id">;
      }
    : never;

There may be confusion as to why TypeScript does not automatically determine that T in the else path (:) of the conditional must belong to the type that was not explicitly checked for in your condition, given that there are only two possible types in the original Union. The explanation lies in TypeScript's limitations when it comes to Control-Flow Analysis of Union Types, particularly in scenarios involving Conditional Types.

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