Question

下面的代码块从一个更大的应用程序中删除，以说明发生了什么...如果我忽略错误，则代码本身可以正常执行。只是类型提示不喜欢它，也不知道为什么。

不过，我确实认为与该Omit类型有关。

出现此错误：

Argument of type '{ avatar: string; height: number; width: number; } & Pick<P & IInputProps, Exclude<keyof P, "firstName" | "lastName" | "avatar">>' is not assignable to parameter of type 'WrappedType<P>'.
    Type '{ avatar: string; height: number; width: number; } & Pick<P & IInputProps, Exclude<keyof P, "firstName" | "lastName" | "avatar">>' is not assignable to type 'P'.

代码是（或gist）：

type Omit<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>;

interface FunctionComponent<P = {}> {
    (props: P, context?: any): string;
}

// Our component props types
interface IInputProps {
    firstName: string;
    lastName: string;
    avatar: string;
}

interface ISubProps {
    width: number;
    height: number;
}

// The helper type that takes an abstract prop type, adds the downstream ISubProps + a type that's based on our IInputProps
type WrappedType<P extends object = {}> = P &
    ISubProps &
    Omit<IInputProps, 'firstName' | 'lastName'>;

type SubComponent<P> = FunctionComponent<WrappedType<P>>;
type WrappedComponent<P> = FunctionComponent<P & IInputProps>;

function factory<P extends object = {}>(
    Component: SubComponent<P>,
): WrappedComponent<P> {

    // The props here are of type P & IInputProps
    return ({ lastName, firstName, avatar, ...rest }) => {
        const restString = Object.entries(rest)
            .map(([key, value]) => `${key}: ${value}`)
            .join('\n');

        // -- THIS BIT DOESNT WORK
        // Component's types are ISubProps + IInputProps (-firstName, -lastName) + P
        const componentResponse = Component(
            {
                avatar,
                height: 10,
                width: 20,
                ...rest,
            },
        );
        // -- TO HERE

        return `FirstName: ${firstName}\nLastName: ${lastName}\n${restString}\n\n--BEGIN--\n${componentResponse}\n--END--`;
    };
}


// Example impl
const test = factory<{ foo: string }>(props => {
    return `hello: ${props.foo}, you have the avatar of ${
        props.avatar
        } with height ${props.height} and width ${props.width}`;
})({
    firstName: 'firstName',
    lastName: 'lastName',
    avatar: 'avatar',
    foo: 'foo',
});

console.log(test);

Answer 1

问题在于Typescript在数学上非常有限，它可以处理包含未绑定类型参数（例如P）的映射和条件类型。

尽管对我们来说似乎很明显，但是编译器无法弄清楚，如果从lastName, firstName, avatar中删除P & { firstName: string; lastName: string; avatar: string; }，则会得到P。只要参数P在其中，编译器就不会尝试解析rest的类型，而是将其余类型键入为Pick<P & IInputProps, Exclude<keyof P, "lastName" | "firstName" | "avatar">>

这里没有安全的方法来帮助编译器，您只需要使用类型断言来让编译器知道rest将是P

const componentResponse = Component({
    avatar,
    height: 10,
    width: 20,
    ...(rest as P),
});

类型的参数不能分配给类型的参数

1 个答案: