typescript - 如何将已知的接口(interface)属性与自定义索引签名相结合？

Question

如何键入一个可以同时拥有几个 的对象声明的可选属性 ，例如:

{ 
    hello?: string, 
    moo?: boolean 
}

以及自定义属性 (必须是函数)，例如:

    [custom: string]: (v?: any) => boolean

这是我想看到的，例如:

const myBasic: Example = {moo: false}
// -> ✅ Valid! Using known keys

const myValid: Example = {hello: 'world', customYo: () => true}
// -> ✅ Valid! "customYo" is a function returning a bool. Good job!

const myInvalid: Example = {hello: 'world', customYo: 'yo!'}
// -> ☠️ Invalid! "customYo" must be a function returning a boolean

尝试将索引签名添加到具有已知键的接口(interface)(即 hello?: string, moo?: boolean )要求所有键都是索引签名类型的子集(在这种情况下，函数返回 boolean )。这显然失败了。

Answer 1

所有者(直到现在)接受的问题是不正确的。
以下是您的操作方法:
您需要使索引签名成为接口(interface)中可以包含的所有类型的联合类型:

interface IExample {
    hello?: string;
    moo?: boolean;
    [custom: string]: string | boolean | YourFunctionType;
}

interface YourFunctionType {
    (v?: any): boolean;
}

请注意，我已将您的函数类型提取到单独的界面中以提高可读性。
影响:
这意味着，明确定义的属性得到 TS 的良好支持:

const test: IExample = <IExample>{};
test.hello.slice(2); // using a string method on a string --> OK
const isHello = test.hello === true; // ERROR (as expected): === cannot be applied to types string and boolean
const isMoo2 = test.moo === true; // OK

然而，索引签名中的所有属性现在都需要使用类型保护来检查，这会增加一点运行时开销:

test.callSomething(); // ERROR: type 'string | boolean | YourFunctionType' has no compatible call signatures
if (typeof test.callSomething === 'function') { // alternatively you can use a user defined type guard, like Lodash's _.isFunction() which looks a little bit nicer
    test.callSomething(); // OK
}

另一方面:运行时开销是必要的，因为它可能是 test像这样访问:

const propertyName: string = 'moo';
test[propertyName](); // ERROR: resolves to a boolean at runtime, not a function ...

// ... so to be sure that an arbitrary propertyName can really be called we need to check:
const propertyName2: string = 'arbitraryPropertyName';
const maybeFunction = test[propertyName2];
if (typeof maybeFunction === 'function') {
    maybeFunction(); // OK
}