Typescript 基于泛型的柯里化函数的条件参数

发布于08月23日

如何使参数只有在没有提供泛型的情况下才有条件？

type Params<T> = T extends Record<string, string> ? T : never;

export const getPath =
  <T>(path: string) =>
  (params: Params<T>) => {
    const lang = getCurrentLang();

    return generatePath(path, { lang, ...(params ?? {}) });
  };

例如:

const login = getPath("/:lang/login")
login() // should works without types error

const book = getPath<{ id: string }>("/:lang/books/:id")
book() // error
book({ id: "1" }) // works

推荐答案

您可以将泛型与条件类型结合使用，以强制getPath()的某个返回类型.让您泛型参数T扩展条件返回值的Union Type和不能赋予该值的值(例如undefined)，这样您就可以判断是否传递了泛型参数.

If a generic argument was passed, return (param: T) => ... otherwise just () => .... This logic can be simplified even more using tuples in rest parameters and spread expressions:
(...params: T extends Record<string, string> ? [T] : [] ) => any

export const getPath = <
  T extends Record<string, string> | undefined = undefined
>(
  path: string
): ((
  ...params: T extends Record<string, string> ? [T] : []
) => any) => {
  return null as any; // ignoring the implemetation
};

const login = getPath("/:lang/login");
login(); // works

const book = getPath<{id: string}>("/:lang/books/:id");
//    ^? const book: (param: {id: string;}) => any
book(); // error
book({id: "1"}); // works

TypeScript Playground个

附注:我删除了函数实现，因为我不知道getCurrentLang()和generatePath()是如何定义的，也不知道它们的返回类型是什么.