Unexpected type inference on array fields

[Eprince-hub]

Describe the bug
SafeQL infers any type on fields that returns an array, leading to a Query has incorrect type annotation error when type any is not passed to this field.
The example below will throw the above error on FoodNames because a type other than any is being passed

type AnimalWithFoodName = {
  firstName: string;
  foodNames: string[];
};

export async function queryJsonAgg() {
  return await sql<AnimalWithFoodName[]>`
    SELECT
       first_name,
       (SELECT json_agg(name) FROM foods) AS food_names
    FROM
      animals
  `;
}

To Reproduce
Steps to reproduce the behavior:

1. Setup SafeQL
2. Use this code in .ts file

type AnimalWithFoodName = {
  firstName: string;
  foodNames: string[];
};

export async function queryJsonAgg() {
  return await sql<AnimalWithFoodName[]>`
    SELECT
       first_name,
       (SELECT json_agg(name) FROM foods) AS food_names
    FROM
      animals
  `;
}

Expected behavior
A clear and concise description of what you expected to happen.
I expect that SafeQL doesn't throw an error when the correct type is being passed and should infer the correct type
Using type any is not encouraged in TypeScript and should also be discouraged in SafeQL

Screenshots
If applicable, add screenshots to help explain your problem.
Incorrect type annotation error when the correct type is passed

No error when any type is passed

Desktop (please complete the following information):

• OS: MAC OS
• PostgreSQL version 15
• Version [e.g. 22]

Additional context
Add any other context about the problem here.

[Newbie012]

json_agg method returns a json, which by default I map to any since the majority of the Typescript community prefer any over unknown. But you could change that behavior in the overrides.types.

What you want is array_agg instead:

type AnimalWithFoodName = {
  firstName: string;
  foodNames: string[];
};

export async function queryJsonAgg() {
  return await sql<AnimalWithFoodName[]>`
    SELECT
       first_name,
       (SELECT array_agg(name) FROM foods) AS food_names
    FROM
      animals
  `;
}

[Eprince-hub]

Thank you, @Newbie012, I will give this a try