Question

我正在使用这些接口

--globalize-symbol

我正在编写一个辅助函数，它给出一个路径，增加集合中实体的相应值。

示例：

/**
 * Interface for entities
 */
export interface Entity extends Object {
    readonly id: EntityId;
}

/**
 * Interface for collection-based state
 */
export interface Collection<T extends Entity> extends Object {
    readonly entities: { [key: string]: T };
    readonly ids: EntityId[];
}

这些是我目前的打字

Collection.increment(myCollection, 'some-id', ['stats', 'total']);

对于这个例子，由于它们习惯于对传入的类型进行检查，所以打字非常冗长

上述类型有效，唯一的问题是非数字类型的路径仍然有效。

    export function increment<
        C extends Collection<any>,
        E = C extends Collection<infer U> ? U : never,
        K1 extends keyof E = keyof E,
        V1 extends E[K1]= E[K1]
        >(collection: C, entityId: string, path: K1 | [K1]): C
    export function increment<
        C extends Collection<any>,
        E = C extends Collection<infer U> ? U : never,
        K1 extends keyof E = keyof E,
        V1 = Exclude<E[K1], void>,
        K2 extends keyof V1 = keyof V1,
        V2 extends V1[K2]= V1[K2]
        >(collection: C, entityId: string, path: K1 | [K1] | [K1, K2]): C

但上述内容并未按预期强制实施约束

有什么想法吗？

Answer 1

您需要按属性类型过滤可用的键。从Omit类型中获取提示，我们可以使用条件类型来创建一个类型，该类型将根据属性类型过滤掉属性。

type FilterKeysByType<T, U> = ({[P in keyof T]: T[P] extends U ? P: never } & { [x: string]: never })[keyof T];  
// If there is a single key, it must be a key of a number field
export function increment<
    C extends Collection<any>,
    E = C extends Collection<infer U> ? U : never,
    K1 extends FilterKeysByType<E, number | undefined> = FilterKeysByType<E, number | undefined>,
    V1 extends Exclude<E[K1], void> = Exclude<E[K1], void>
>(collection: C, entityId: string, path: K1 | [K1]): C
// If there are two keys, the first key, can be any  type, but the second key must be the key of a number field 
export function increment<
    C extends Collection<any>,
    E = C extends Collection<infer U> ? U : never,
    K1 extends keyof E = keyof E,
    V1 = Exclude<E[K1], void>,
    K2 extends FilterKeysByType<V1, number| undefined> = FilterKeysByType<V1, number| undefined>,
    V2 extends V1[K2]= V1[K2]
    >(collection: C, entityId: string, path: [K1, K2]): C

// Usage
declare var c : Collection<Comment>;

interface Comment extends Entity{
    name: string;
    value: number;
    optValue?: number;
    subcomment: Comment;
    optSubcomment?: Comment;
}

increment(c, "", "value"); //ok
increment(c, "", "optValue"); //ok
increment(c, "", "name");// error
increment(c, "", ["subcomment", "value"]); // ok
increment(c, "", ["subcomment", "optValue"]); // ok
increment(c, "", ["subcomment", "name"]); // error
increment(c, "", ["optSubcomment", "value"]); // ok
increment(c, "", ["optSubcomment", "optValue"]); // ok
increment(c, "", ["optSubcomment", "name"]); // error

Answer 2

我尽量不打破条件类型，除非我没有它们就无法解决问题（因为它们有时会有幽灵般的推理属性，我无法将我的大脑包裹起来）。那么这些打字怎么样呢？

declare function increment<
 C extends Collection<Entity & Record<K1,Record<K2,number>>>, 
 K1 extends string, 
 K2 extends string>(collection: C, entityId: string, path: [K1, K2]): C;
declare function increment<
 C extends Collection<Entity & Record<K1, number>>, 
 K1 extends string>(collection: C, entityId: string, path: K1 | [K1]): C;

只有C和K1（可能是K2）。我的想法是C应该是Collection<Entity & Something>，其中Something是Record<K1, number>或Record<K1, Record<K2, number>>，具体取决于{{1}中的元素数量}}。（如果重要，path定义为Record<K extends string, V>，它只是意味着“具有{[P in K]: V}类型的键和类型K的值”的对象。）

您可以验证其行为符合要求：

希望有所帮助;祝你好运！

强制类型是接口的子类型并扩展数量

2 个答案: