typescript-- How do you type generics?

[RyanDurk]

I wrote some example code below with the issue. Basically the table has an id, an "inner" jsonb column and a "type" column.

The error is on the very last line. My workaround is just suffixing an "as T[]" Is there a better way?

import sql from "./server/db"

type MyType = "child" | "other"

interface Parent {
  id: number
  inner: object
  type: MyType
}

interface Child extends Parent {
  type: "child"
  inner: {
    name: string
  }
}

async function run1(): Promise<Child[]> {
  const x = await sql<Child[]>`select id, inner from table`
  return x
}

run1()

async function run2<T extends Parent>(type: MyType): Promise<T[]> {
  const x = await sql<T[]>`select id, inner from table where type = ${type}`
  return x
}

gives this error:

  The types returned by 'pop()' are incompatible between these types.
    Type 'TransformRow<T> | undefined' is not assignable to type 'T | undefined'.
      Type 'TransformRow<T>' is not assignable to type 'T | undefined'.
        Type 'T | { '?column?': T; }' is not assignable to type 'T | undefined'.
          Type '{ '?column?': T; }' is not assignable to type 'T'.
            'T' could be instantiated with an arbitrary type which could be unrelated to '{ '?column?': T; }'.ts(2322)```

[Minigugus]

Short answer: indeed, it's a bug... 😞

Long answer: historically, if was possible to pass types like number as row type, and these would be become { '?column?': T }, as postgresql does with unamed columns (like in SELECT 42;). But it was too painful and not really used so contraints on T where added, to enforce objects only (see types), then transforming to '?column?' isn't even possible anymore, so that part should have been removed, to avoid bugs like in your case 😅

A fix is on the way: #383