这是一个很长的机会,但我希望能够从现有的接口生成一个新的接口,但使用为原始接口派生的键名。
下面是我想做的事情的一个例子:
给定此源对象:
interface Car {
brand: Brand;
model: string;
year: number;
}
我希望能够声明一个接口:
interface CarSetters {
setBrand: (brand: Brand) => any;
setModel: (model: string) => any;
setYear: (year: number) => any;
}
但我不想像上面那样手动声明它。相反,我希望能够像这样使用它:
type CarSetters = Setters<Car>
或
type CarState = Car & Setters<Car>
它看起来是一个非常简单的映射:
type Brand = boolean;
interface Car {
brand: Brand;
model: string;
year: number;
};
type CarSetters = {
[T in keyof Car as `set${Capitalize<T>}`]: (newValue: Car[T]) => any;
};
为了防止符号引起问题,您可以执行以下操作:
type CarSetters = {
[T in keyof Car as T extends symbol ? never : `set${Capitalize<T>}`]: (newValue: Car[T]) => any;
};
或
type CarSetters = {
[T in keyof Car as T extends symbol ? never : `set${Capitalize<T>}`]: (newValue: Car[T]) => any;
} & {
[T in keyof Car as T extends symbol ? T : never]: Car[T];
};
我能够找到一个符合我要求的泛型,这主要归功于CertainPerformance的回答,我已经接受了它,因为它确实明确地回答了我提出的问题。但我也在分享我想出的仿制药:
type Setters<T> = {
[K in keyof T as K extends string ? `set${Capitalize<K>}` : never]: (
value: T[K]
) => any;
};
type State<T> = T & Setters<T>