Object Key Paths

题目来源(type-challenges)

问题 #

实现一个 _.get 拿到所有索引的路径

const ref = {
  count: 1,
  person: {
    name: 'cattchen',
    age: 22,
    books: ['book1', 'book2'],
    pets: [
      {
        type: 'cat',
      },
    ],
  },
}
ObjectKeyPaths<{ name: string, age: number }> // 'name' | 'age'
ObjectKeyPaths<{
  refCount: number
  person: { name: string, age: number }
}>,
// 'refCount' | 'person' | 'person.name' | 'person.age'
ObjectKeyPaths<typeof ref> // 'count'
ObjectKeyPaths<typeof ref> // 'person'
ObjectKeyPaths<typeof ref> // 'person.name'
ObjectKeyPaths<typeof ref> // 'person.age'
ObjectKeyPaths<typeof ref> // 'person.books'
ObjectKeyPaths<typeof ref> // 'person.pets'
ObjectKeyPaths<typeof ref> // 'person.books.0'
ObjectKeyPaths<typeof ref> // 'person.books.1'
ObjectKeyPaths<typeof ref> // 'person.books[0]'
ObjectKeyPaths<typeof ref> // 'person.books.[0]'
ObjectKeyPaths<typeof ref> // 'person.pets.0.type'
ObjectKeyPaths<typeof ref> // person.notExist -> never

解答 #

type IsNumber<T> = T extends number ? `[${T}]` | `.[${T}]` : never

type ObjectKeyPaths<
  T extends object,
  Flag extends boolean = false,
  K extends keyof T = keyof T
> = K extends string | number
  ? (Flag extends true ? `.${K}` | IsNumber<K> : `${K}`) |
  (
    T[K] extends Record<string, any>
      ? `${Flag extends true ? `.${K}` | IsNumber<K> : `${K}`}${ObjectKeyPaths<T[K], true>}`
      : never
  )
  : never

拆分 #

• 判断当前索引是否为字符串或数字
• 如果不是则返回 never
• 如果是则根据是否是第一层返回不同的结果的 key
• 并且联合上 K 索引的值
• 如果是对象则继续递归联合