在 Typescript 中，如何允许类型具有指向该类型递归值的泛型的所有键？

Question

Right now I have something like this where "any" can act as the generic "V"

interface Validation<V> {
  $isValid: boolean
  $isValidating: boolean
  $value: V
  [prop: string]: boolean | V | Validation<V>
}

What I'd like to do is replace the string indexing type with any string key K from V will return a sub Validation interface.

interface Validation<V> {
  $isValid: boolean
  $isValidating: boolean
  $value: V
  [K extends Extract<keyof V, string>]: Validation<V[K]>
}

This clearly doesn't work and curious to know if something similar is achievable.

Answer 1

Yes. Here is how to do it

type Validation<V> = {
    [K in Extract<keyof V, string>]: Validation<V[K]>
} & {
    $isValid: boolean
    $isValidating: boolean
    $value: V
}

1. You should use mapped type , so you should write K in ... .

   The syntax resembles the syntax for index signatures with a for .. in inside. There are three parts:

   • The type variable K, which gets bound to each property in turn.

   • The string literal union Keys, which contains the names of properties to iterate over.

   • The resulting type of the property.

2. To dd additional properties use intersection , as it is not possible to add additional properties to mapped type directly.