具有映射和条件类型的递归类型定义

Question

我正在尝试想出一种方法来使用 TypeORM 获得更好的类型安全性。以下是一些 TypeORM 实体定义示例。

import { BaseEntity, Entity, Column, ManyToMany, JoinTable, ManyToOne, OneToMany } from 'typeorm';

@Entity()
class Product extends BaseEntity {
  @Column({ type: 'text' })
  public name: string;

  @Column({ type: 'text' })
  public description: string;

  @ManyToMany(_ => Category, category => category.products)
  @JoinTable()
  public categories: Category[];
}

@Entity()
class Category extends BaseEntity {
  @Column({ type: 'text' })
  public name: string;

  @ManyToMany(_ => Product, product => product.categories)
  public products: Product[];

  @ManyToOne(_ => Supplier, supplier => supplier.categories, { nullable: false })
  public supplier: Supplier;
}

@Entity()
class Supplier extends BaseEntity {
  @Column('text')
  public name: string;

  @Column({ type: 'boolean', default: true })
  public isActive: boolean;

  @OneToMany(_ => Category, category => category.supplier)
  public categories: Category[];
}

我正在尝试定义一种类型，该类型仅对作为实体本身的实体的属性有效。这最好用一个例子来解释：

type Relations<T extends BaseEntity> = {
  // An object whose:
  // - Keys are some (or all) of the keys in type T, whose type is something which extends BaseEntity.
  // - Values are another Relations object for that key.
}

// Some examples

// Type error: "color" is not a property of Product.
const a: Relations<Product> = {
  color: {}
}

// Type error: "name" property of Product is not something that extends "BaseEntity".
const a: Relations<Product> = {
  name: {}
}

// OK
const a: Relations<Product> = {
  categories: {}
}

// Type error: number is not assignable to Relations<Category>
const a: Relations<Product> = {
  categories: 42
}

// Type error: "description" is not a property of Category.
const a: Relations<Product> = {
  categories: {
    description: {}
  }
}

// Type error: "name" property of Category is not something that extends "BaseEntity".
const a: Relations<Product> = {
  categories: {
    name: {}
  }
}

// OK
const a: Relations<Product> = {
  categories: {
    supplier: {}
  }
}

// Type error: Date is not assignable to Relations<Supplier>
const a: Relations<Product> = {
  categories: {
    supplier: new Date()
  }
}

// etc.

到目前为止我想出了以下方法，但它不起作用，甚至可能离正确答案还差得远：

type Flatten<T> = T extends Array<infer I> ? I : T;

type ExcludeNonEntity<T> = T extends BaseEntity | Array<BaseEntity> ? Flatten<T> : never;

type Relations<T extends BaseEntity> = {
  [P in keyof T as ExcludeNonEntity<P>]: Relations<T[P]>;
};

Answer 1

我的建议是这样的：

type DrillDownToEntity<T> = T extends BaseEntity ?
    T : T extends ReadonlyArray<infer U> ? DrillDownToEntity<U> : never;

type Relations<T extends BaseEntity> =
    { [K in keyof T]?: Relations<DrillDownToEntity<T[K]>> }

DrillDownToEntity<T> 类似于您的 Flatten<T> 类型与 ExcludeNonEntity<T> 的混合类型，只是它以递归方式运行。它为任意数量的嵌套提取所有数组元素类型，仅保留可分配给 BaseEntity 的那些类型。观察：

type DrillTest = DrillDownToEntity<Category | string | Product[] | Supplier[][][][][]>
// type DrillTest = Category | Product | Supplier

我不知道你是否会拥有数组的数组；如果不是，则可以使其成为非递归的。但重要的是，任何最终不能分配给 BaseEntity 的类型都会被丢弃。

那么Relations<T>是一个所有可选属性的类型，其键来自T，其值是T的属性的Relations<DrillDownToEntity<>>。一般来说，大多数属性都是 never 类型，因为大多数属性本身不能分配给 BaseEntity。观察：

type RelationsProduct = Relations<Product>;
/* type RelationsProduct = {
    name?: undefined;
    description?: undefined;
    categories?: Relations<Category> | undefined;
    hasId?: undefined;
    save?: undefined;
    remove?: undefined;
    softRemove?: undefined;
    recover?: undefined;
    reload?: undefined;
} */

请注意，never 类型的可选属性和 undefined 类型之一是相同的，至少没有 the --exactOptionalPropertyTypes compiler flag enabled. This has the effect of preventing you from assigning any property of these types at all unless they are undefined. I find this is probably better than merely omitting those properties; a value of type {categories?: Relations<Category>} might or might not have a string-valued name property, according to structural typing，而 {categories?: Relations<Category>, name?: never} 绝对不会有定义的 name 属性。

您可以使用 Relations.

的定义验证您的示例代码是否按预期工作

以下代码：

type Relations<T extends BaseEntity> = {
    [P in keyof T as ExcludeNonEntity<P>]: Relations<T[P]>;
};

不工作有几个原因，其中最直接的是你正在使用 key remapping syntax 大概抑制非 BaseEntity 可分配的属性，但你正在写 ExcludeNonEntity<P>其中 P 是 key 类型。并且没有 keys 会是 BaseEntity，所以这很可能最终排除所有键，即使你可以让它工作。如果你想禁止按键，那么你需要检查 T[P] 而不是 P，然后基于此省略或包含 P。还有其他一些小问题（例如，属性不是可选的），但最大的问题是将键视为值。

Playground link to code

Answer 2

对的改进：

type Relations<T extends BaseEntity> = {
    [K in keyof T as DrillDownToEntity<T[K]> extends never ? never : K]?: Relations<DrillDownToEntity<T[K]>>
}

这样，以下内容也会导致类型错误，而 :

不会导致错误

const b: Relations<Product> = {
    name: undefined
}

此答案的灵感来自 Anders Hejlsberg's PR 中的以下部分：

When the type specified in an as clause resolves to never, no property is generated for that key. Thus, an as clause can be used as a filter:
type Methods<T> = { [P in keyof T as T[P] extends Function ? P : never]: T[P] };
type T60 = Methods<{ foo(): number, bar: boolean }>;  // { foo(): number }

具有映射和条件类型的递归类型定义

Recursive type definition with mapped and conditional types

typescript

typeorm

keyof

mapped-types

conditional-types