How to use TypeScript's indexed access types with nullable nested types?

user_9835872 0
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I'm trying to define a TypeScript type in terms of another type.

This works:

type Result = { data: { nestedData: { foo: string; bar: string } } };

type NestedData = Result['data']['nestedData'];

But, when the data property is nullable, this doesn't work:

type Result = { data: { nestedData: { foo: string; bar: string } } | null };

type NestedData = Result['data']['nestedData'];

and results in the error:

Property 'nestedData' does not exist on type '{ nestedData: { foo: string; bar: string; }; } | null'.(2339)

How can I define my NestedData type in terms of Result, without duplicating any part of Result's typings?

Demo on TypeScript Playground

Edit: I'm getting my NestedData type from a codegen tool and I'm defining NestedData as a shorter type alias. In reality the typing is longer so I want to minimize repetition.

nullability typescript
2 Answers

You can use Exclude to remove null from the union:

type NestedData = Exclude<Result['data'], null>['nestedData']

Playground Link

It makes sense to do it that way when you can't change your Result type for some reason. In other situations it would be more natural to define them this way round:

type NestedData = { foo: string; bar: string }
type Result = { data: { nestedData: NestedData } | null }
user_122990000

Generic util for getting nullable properties:

type Result = {
    data: {
        nestedData: { foo: string; bar: string }
    } | null
};

type GetNullable<T, Prop extends keyof NonNullable<T>> = NonNullable<T>[Prop]

type NestedData = GetNullable<Result['data'], 'nestedData'>

Playground

user_84952540
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