How can I check if an array contains null values inside typescript?

Question

I have an array that might include null or undefined values

const uncertainObjectIds = ['5fc657037ed3e03d506e1e25', null, '5fc657037ed3e03d506e1e26', undefined]

If uncertainObjectIds does include any null or undefined values I will throw an error. However, if it does not and I am sure that the array is full of defined values and no undefined values I wanna use it.

  const doNullValuesExistInsideObjectIds = uncertainObjectIds.some(
    (objectIdOrNull) => !objectIdOrNull,
  );

  if (doNullValuesExistInsideObjectIds) {
    throw new UserInputError(
      'All Users must have ObjectIds',
    );
  }

const usersCollection = await getCollection('users');
  const originalUsers = await usersCollection.find({
    _id: { $in: uncertainObjectIds },
  });

but TypesSript gives me an error.

Type '(ObjectId | null | undefined)[]' is not assignable to type 'ObjectId[]'

What is the best way to handle this use case inside TypeScript?

They are an array of MongoDB ObjectIds, but I will add the quotes. — ATheCoder
– ATheCoder, Commented Dec 1, 2020 at 14:53

Aplet123 · Accepted Answer · 2020-12-01 14:56:24Z

1

You can use a type guard to accomplish this:

function nonNull<T>(arr: (T | null | undefined)[]): arr is T[] {
    return arr.every(v => v !== null && v !== undefined);
}

const mixed = [1, 2, 3, null, 4, 5, undefined, 6];

// uncommenting this line will error due to having null values
// const x: number[] = mixed;

// apply the type guard
if (!nonNull(mixed)) {
    throw new Error("Cannot have null values.");
}

// no longer has null values
const y: number[] = mixed;

answered Dec 1, 2020 at 14:56

Aplet123

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