TypeScript 怎么用递归类型实现深层路径类型推导？

Question

仓乃文 · Accepted Answer

访问深层嵌套对象属性时，路径字符串如 "user.address.city" 没有类型校验，拼写错误不会报错。递归类型可以推导出对象所有合法的深层路径，实现类型安全的路径访问。推导所有路径： ```typescript type Path = T extends object ? { [K in keyof T & string]: K | `${K}.${Path}` }[keyof T & string] : never; interface State { user: { name: string; address: { city: string; zip: string } }; settings: { theme: string; lang: string }; } type StatePath = Path; // "user" | "user.name" | "user.address" | "user.address.city" | "user.address.zip" // | "settings" | "settings.theme" | "settings.lang" ``` 类型安全的路径访问： ```typescript function getByPath>(obj: T, path: P): any { return path.split(".").reduce((acc: any, key) => acc[key], obj); } const state: State = { user: { name: "Tom", address: { city: "BJ", zip: "100" } }, settings: { theme: "dark", lang: "zh" } }; getByPath(state, "user.address.city"); // OK getByPath(state, "user.address.foo"); // 报错：路径不存在 ``` 补充说明： - 递归深度受 TypeScript 限制，通常 5-7 层以内安全 - & string 约束键为字符串类型，排除 symbol 键 - 复杂递归类型可能影响编译性能 - 适合状态管理库（如 immer、lodash.get 的类型安全版） - TS 4.1+ 模板字面量类型使路径推导成为可能