從 Typescript 界面中選擇鍵並更改它們的類型

Question

我有這個界面：

export interface MutationSheetDefinition {
    _type: "mutation.sheet.definition"
    id: string
    name: string
    mutations: MutationDefinition[]
    referencedMutationSheets: MutationSheetReference[]
}

我想要這種類型：

interface ValidatedMutationSheetDefinition {
    name: ValidationOf<string>
    mutations: ValidationOf<MutationDefinition>[]
    referencedMutationSheets: ValidationOf<MutationSheetReference>[]
}

我知道我可以使用以下方法選擇我們的界面子集：

type narrowed = Pick<MutationSheetDefinition, "name"| "mutations" | "referencedMutationSheets">

這會給我

interface {
    name: string
    mutations: MutationDefinition[]
    referencedMutationSheets: MutationSheetReference[]
}

但是我可以更改拾取鍵的類型嗎？

我會為所有選擇的鍵選擇一個固定類型，所以這樣做：

interface {
    name: Validation
    mutations: Validation[]
    referencedMutationSheets: Validation[]
}

Answer 1

你可以這樣做：

type PickAndChangeToValidationOf<T, P extends keyof T> = {
  [K in P]: T[K] extends (infer I)[] 
    ? ValidationOf<I>[]
    : ValidationOf<T[K]> 
}

我不確定您希望如何處理數組。 在您的示例中，您刪除了[]並將其放在ValidationOf<T>后面。 上面的類型也這樣做，但僅適用於第一級數組。 如果您希望雙嵌套數組具有不同的行為，則應指定。

用法：

interface MutationSheetDefinition {
    _type: "mutation.sheet.definition"
    id: string
    name: string
    mutations: number[]
    referencedMutationSheets: number[]
}

type Narrowed = PickAndChangeToValidationOf<MutationSheetDefinition, "name"| "mutations" | "referencedMutationSheets">
// type Narrowed = {
//     name: ValidationOf<string>;
//     mutations: ValidationOf<number>[];
//     referencedMutationSheets: ValidationOf<number>[];
// }

操場

Answer 2

你在那里（我用隨機字符串交換了你的自定義類型，只需刪除它們）：

type MutationDefinition = "MutationDefinition";
type MutationSheetReference = "MutationSheetReference";

export interface MutationSheetDefinition {
  _type: "mutation.sheet.definition"
  id: string
  name: string
  mutations: MutationDefinition[]
  referencedMutationSheets: MutationSheetReference[]
}

type Narrowed = Pick<MutationSheetDefinition, "name"| "mutations" | "referencedMutationSheets">;

type Validation = "Validation";

type Validational<Type> = {
  [Property in keyof Type]: Type[Property] extends unknown[] ? Validation[] : Validation;
};

type Definition = Validational<Narrowed>;

const d: Definition = {
  name: "Validation",
  mutations: ["Validation", "Validation"],
  referencedMutationSheets: ["Validation", "Validation"],
}

此解決方案的一個缺點是它的類型硬編碼為僅數組和所有其他內容。 如果您使用任何其他類型，則必須添加更多類型，即需要添加數組數組。