CODEWITHSUNDEEP
typescripttypeguards
Rico Kahler
Rico Kahler

Reputation: 19202

Typescript user-defined type guard checking object has all properties in array

I'm trying to write a user-defined type guard that tests whether the value it's given has all the properties in a given array.

I'm calling this function hasAll and it's implementation and usage in Javascript would look like this:

function hasAll(obj, keysToCheck) {
  if (!obj) return false;

  for (const key of keysToCheck) {
    const value = obj[key];
    if (value === null) return false;
    if (value === undefined) return false;
  }

  return true;
}

hasAll({ foo: 'test', bar: 5 }, ['foo', 'bar']); // true

hasAll({ foo: 'test', bar: 5 }, ['foo', 'bar', 'baz']); // false

What I'm trying to do now is turn the above function into a type guard. This is what I have so far:

// this _almost_ works 🔴
type Nullable<T> = T | null | undefined;

type RemoveNullables<T, K extends keyof T> = {
  [P in K]-?: T[P] extends Nullable<infer U> ? U : T[P];
};

function hasAll<T, K extends keyof NonNullable<T>>(
  obj: T,
  keysToCheck: K[],
): obj is RemoveNullables<NonNullable<T>, K> {
  // but i'm getting an error here 👆👆👆
  if (!obj) return false;

  const nonNullableObj = obj as NonNullable<T>;

  for (const key of keysToCheck) {
    const value = nonNullableObj[key];
    if (value === null) return false;
    if (value === undefined) return false;
  }

  return true;
}

export default hasAll;

playground link

The error message is:

A type predicate's type must be assignable to its parameter's type.
  Type 'RemoveNullables<NonNullable<T>, K>' is not assignable to type 'T'.

I've read this answer with a good explanation however it doesn't really help my case.

I want to explicitly assert that my type T will conform to RemoveNullables<NonNullable<T>, K> after it runs through this function. I don't really care whether or not T is assignable to RemoveNullables<NonNullable<T>, K> (if that makes sense).

  1. Am I going about this wrong? Is there a better way to write this type guard?
  2. If this approach is fine, how can I tell typescript that I don't care if the type guard is "unsafe" per se?

Upvotes: 3

Views: 2190

Answers (1)

kaya3
kaya3

Reputation: 51034

This seems to meet your requirements:

type ExcludeNullable<T, K extends keyof NonNullable<T>> = NonNullable<T> & {
    [k in K]-?: Exclude<NonNullable<T>[k], null | undefined>
}

function hasAll<T, K extends keyof NonNullable<T>>(
    obj: T,
    keysToCheck: K[]
): obj is ExcludeNullable<T, K> {
    return obj !== null && obj !== undefined
        && keysToCheck.every(k => obj![k] !== null && obj![k] !== undefined);
}

A few notes:

  • The T & ... intersection type guarantees that ExcludeNullable<T, K> is assignable to T. Without this, the mapped type doesn't have properties of T that are missing from K.
  • Exclude is a simpler way to get rid of null and undefined than using a conditional type with infer.
  • I took the liberty of simplifying the hasAll function implementation a bit.

Playground Link

Upvotes: 5

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