使用 Typescript 时有没有办法推断映射类型的键？

Question

我有一个代表我的业务运营的数据结构：

const operations = {
    list: {
        request: "a",
        response: "b",
    },
    get: {
        request: "a",
        response: "b",
    },
} as const;

我想创建一个基于上面的

operations

接受回调的函数：

type Ops = typeof operations;

type Callbacks = {
    [Property in keyof Ops]: (
        param: Ops[Property]["request"]
    ) => Ops[Property]["response"];
};

现在，如果我想定义我的回调，如果我错过其中任何一个，编译器会抱怨：

const callbacks: Callbacks = {

};
// ^^^--- Type '{}' is missing the following properties from type 'Callbacks': list, get

现在我的问题是我想创建另一种类型，以便我可以键入检查

operations

对象的结构：

interface OperationDescriptor<A, B> {
    request: A;
    response: B;
}

type Operations = {
    [key: string]: OperationDescriptor<any, any>;
};

const operations: Operations = {
    list: {
        request: string,
        response: string,
    },
    get: {
        request: string,
        response: string,
    },
} as const;

但是当我这样做时，编译器将不再抱怨，因为

Operations

不知道我在

operations

中的键。有没有办法鱼和熊掌兼得，例如：

• 定义一个类型，我可以用它来检查

  operations

  和
的结构
• 有一个

  Callbacks

  类型，它将根据

  operations

  的结构对我的回调函数进行类型检查？

Answer 1

您可以使用

satisfies

运算符来检查值是否可分配给某个类型，而无需将其扩大到该类型。它正是针对类型注释会忘记您关心的信息的情况而设计的：

const operations = {
    list: {
        request: "a",
        response: "b",
    },
    get: {
        request: "a",
        response: "b",
    },
} as const satisfies Operations;

如果以上编译通过，那么你就知道你没问题。否则你会得到你期望的错误：

const badOperations = {
    list: {
        request: "a",
        response: "b",
    },
    get: { // error!
    //~ <-- Property 'response' is missing
        request: "a",
    },
} as const satisfies Operations;

类型

typeof operations

仍然与您需要的一样详细：

type Ops = typeof operations;
/* type Ops = {
    readonly list: {
        readonly request: "a";
        readonly response: "b";
    };
    readonly get: {
        readonly request: "a";
        readonly response: "b";
    };
} */

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