还原机工厂打印问题中的Typescirpt映射

Question

我正在尝试实现类型安全功能，以创建基于“动作处理程序”映射的还原器。这样做的目的是让API看起来像这样：

export const Actions = {
    setToken: (token: string) => createAction(SET_TOKEN_TYPE, token),
    invalidateToken: () => createAction(INVALIDATE_TOKEN_TYPE),
    startLogin: () => createAction(START_LOGIN_TYPE)
};

export const reducer = createReducer<State, typeof Actions>(
    {
        [SET_TOKEN_TYPE]: ({ loginError, ...state }, action) => ({
            ...state,
            token: action.payload,
            loading: false
        }),
        [INVALIDATE_TOKEN_TYPE]: ({ token, ...state }) => state,
        [START_LOGIN_TYPE]: ({ loginError, ...state }) => ({
            ...state,
            loading: true
        })
    },
    {
        loading: false
    }
);

createReducer函数(为了清晰起见没有类型记录)应该如下所示：

function createReducer(handlers, initialState) {
    return (state = initialState, action) => {
        if (action.type in handlers) {
            return handlers[action.type](state, action);
        }
        return state;
    };
}

我创建了这样的类型化函数以具有类型安全性：

interface Action<T extends string> {
    type: T;
}
type ActionCreator<T extends string> = (...args: any) => Action<T>;
type ActionsCreators = {
    [creator: string]: ActionCreator<any>;
};
type ActionsUnion<Actions extends ActionsCreators> = ReturnType<
    Actions[keyof Actions]
>;
type ActionHandlers<ActionCreators extends ActionsCreators, State> = {
    [K in ReturnType<ActionCreators[keyof ActionCreators]>["type"]]: (
        state: State,
        action: ReturnType<ActionCreators[K]> 
    ) => State
};

    function createReducer<State, Actions extends ActionsCreators>(
    handlers: ActionHandlers<Actions, State>,
    initialState: State
) {
    return (
        state: State = initialState,
        action: ActionsUnion<Actions>
    ): State => {
        if (action.type in handlers) {
            // unfortunately action.type is here any :(
            return handlers[action.type](state, action); // here I have the error
        }
        return state;
    };
}

在handlers[action.type]中，我有错误(使用noImplicitAny: true)

元素隐式具有“任意”类型，因为类型'ActionHandlers‘没有索引签名。

知道怎么在减速机内输入action.type吗？

您可以在要旨中找到整个示例

Answer 1

您得到错误的原因是action.type被隐式地键入为any，因为没有应用适用的类型。

在链的某一点上，您使用any作为需要根植于string的类型参数。

type ActionsCreators = {
    [creator: string]: ActionCreator<any>;
};

如果在这里添加一个类型参数，您可以替换any；但是，您需要一直传递它。

请参阅进行此更新的以下版本。我不得不将一些中间类型重命名为泛型名称(T或P)，因为我很难保持输入的正确性。

使用额外的类型参数<P>，我们现在有一个类型用于下面的内容，而不是隐式any

const f = handlers[action.type];

在这里，f变成了ActionHandlers<P, T, State>[P]

export interface Action<T extends string> {
    type: T;
}

export interface ActionWithPayload<T extends string, P> extends Action<T> {
    payload: P;
}

export type ActionCreator<T extends string> = (...args: any) => Action<T>;

export type ActionsCreators<T extends string> = {
    [creator: string]: ActionCreator<T>;
};

export type ActionsUnion<P extends string, T extends ActionsCreators<P>> = ReturnType<T[keyof T]>;

export type ActionHandlers<P extends string, T extends ActionsCreators<P>, State> = {
    [K in ReturnType<T[keyof T]>["type"]]: (
        state: State,
        action: ReturnType<T[K]>
    ) => State
};

export function createReducer<P extends string, State, T extends ActionsCreators<P>>(
    handlers: ActionHandlers<P, T, State>,
    initialState: State
) {
    return (state: State = initialState, action: ActionsUnion<P, T>): State => {

        if (action.type in handlers) {
            const f = handlers[action.type];
            return f(state, action);
        }

        return state;
    };
}