如何在对象（Typescript）中强制执行一致的类型？

Question

我已经做到了这一点：这似乎有效

function test<types extends Record<string,any>>(dict: dictionary<types>){}

type dictionary<types extends Record<string, any>> = {
  [key in keyof types]: {
    bar?: types[key];
    foo?: (value:types[key])=>true;
  };
};

test({
 key1:{
  bar: 2,
  foo: (input:number)=>true,
 },
 key2:{
  bar: 'hello'
  foo: (input: number)=>true, // Error! "input" needs to be string
 }
})

但是！ 我还需要对dict 参数的泛型类型引用。而且由于某种原因，这不起作用

function test2<
  types extends Record<string,any>,
  dictionary extends dictionary2<types> // <-- Added a generic type
>(dict: dictionary){}

// Same as above
type dictionary2<types extends Record<string, any>> = {
  [key in keyof types]: {
    bar?: types[key];
    foo?: (value:types[key])=>true;
  };
};

// Same as above
test2({
 key1:{
  bar: 2,
  foo: (input: number)=>true,
 },
 key2:{
  bar: 'hello', 
  foo: (input:number)=>true,// Should be an Error (but isn't)! "input" needs to be string
 }

Playground link

Answer 1

你可以这样做：

function test2<T extends Record<string, unknown>>(dict: Dictionary<T>) { }

type Dictionary<T> = {
  [key in keyof T]: {
    bar?: T[key];
    foo?: (value: T[key]) => true;
  };
}

// Same as above
test2({
  key1: {
    bar: 2,
    foo: (input: number) => true,
  },
  key2: {
    bar: 'hello',
    foo: (input: number) => true, // Actual error
  }
});

TypeScript playground

Answer 2

更改推理范围，以便推断 types 并根据该推理而不是第二个类型参数键入 dict，即，

function test2<
  types extends Record<string,any>
>(dict: dictionary2<types>){}

工作场所here.

编辑：常规大写的用法示例

function test2<
  T,
>(dict: Dictionary<T>){

  type FullDictionary = Dictionary<T> // can't you just use this in your function?

}

type Dictionary<T extends Record<string, any>> = {
  [K in keyof T]: DictionaryEntry<T[K]>
}

type DictionaryEntry<T> = {
  bar?: T
  foo?: (value:T)=>true
}

test2({
 key1:{
  bar: 2,
  foo: (input: number)=>true,
 },
 key2:{
  bar: 'hello', 
  foo: (input:number)=>true
 }
})