Reputation: 672
I have the following code:
const arrayFromForm: IFirst[] | ISecond[] | IThird[]; // = ... it's initialized
const arrayDB: IFirst[] | ISecond[] | IThird[]; // = ... it's initialized
const entitiesFromDb = arrayDB.filter(
(entity: IFirst | ISecond | IThird) =>
!(arrayFromForm.map(({ id }) => id))
.includes(entity.id)
);
And it gives me 1 error: "The map expression is not callable"
.
Correct me if I'm wrong, but I think that this is because it doesn't know that there is a property named id
in all of the interfaces in the union.
I've used a type guard before, but I think you can create a type guard only between 2 types in an union, because you have to return a boolean?
I want to solve this problem with typescript (user-defined type guard, typeof, instanceof, etc.), but I'm not sure that this is possible, isn't it? instanceof
type guards can't be created for interfaces, right?
What's a good solution here?
EDIT: I've just solved the problem with an interface with the properties that the interfaces in the union have in common and I just changed the type union with the new interface. This is not very good solution as I would want to use specific methods/properties in the different interfaces that I would've had in the type union.
So I'm not deleting the question, because it's still helpful to know if there is a such thing as an user-defined type guards for type unions with more than 2 types in them.
Upvotes: 2
Views: 376
Reputation: 611
This seems like a known issue in Typescript. See Here.
But for now you can do this:
const arrayFromForm: IFirst[] | ISecond[] | IThird[]; // = ... it's initialized
const entitiesFromDb = arrayDB.filter(
(entity: IFirst | ISecond | IThird) =>
!(arrayFromForm as {id: number}[]).map(item => item.id).includes(entity.id) // Pay attention to the brackets
);
We cast the arrayFromForm
as just one appropriate type, instead of a union of types. This is close to the solution you came up with but I think it's easier to maintain it this way.
Upvotes: 1
Reputation: 10979
Try this :-
const elementsIds = arrayFromForm.filter(arrItem => arrItem.id).map((item)=>item.id);
const entitiesFromDb = arrayDB.filter(
(entity: IFirst | ISecond | IThird) =>
!elementsIds.includes(entity.id)
)
Upvotes: 0