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我有三个服务，它们根据不同的操作返回相同类型的对象。我怎样才能以干净的分离来保存特征存储中的所有对象。

行动 1：LOAD_CANDIDATES



    I have an effect that invokes below service call

    public getCandidates(): Observable {
        const url = 'Candidates Url';
        return this.http.get(url);
    }

行动 2：LOAD_MATCHED_CANDIDATES



    I have an effect that invokes below service call

    public getMatchingCandidates(conditions: any): Observable 
    {
        const url = 'Matched Candidates Url';
        return this.http.get(url);
    }

行动 3：LOAD_ASSIGNED_CANDIDATES



    I have an effect that invokes below service call

    public getAssignedCandidates(id: number): Observable {
        const url = 'Assigned candidates url';
        return this.http.get(url);
    }

我确实对他们每个人都有成功和失败的影响。



    Candidate reducer :

    export const reducers = {
      search: fromSearch.reducer,
      candidates: fromCandidates.reducer,
      collection: fromCollection.reducer
    };

    Here is the injection of feature store to module

    StoreModule.forFeature('candidates', combineReducers(fromStore.reducers))

我怎样才能在特征存储中同时拥有所有匹配和分配的候选对象，并通过明确的分离来指定相应的对象类型（即这些是匹配的，这些是分配的）

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1 回答 1

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为了实现这一点，您需要使用多个属性对相应的状态进行切片，并在 reducer 中进行处理。请参阅下面的代码以获取示例实现

将“候选人”声明为“CandidateState”类型

export interface CandidateState {
  allcandidates: Array<Object>;
  matchedcandidates: Array<Object>;
  assignedcandidates: Array<Object>;
}

现在组成你的候选减速器如下

export const candidateInitialState: fromState.CandidateState = {
  allcandidates: [],
  matchedcandidates: [],
  assignedcandidates: []
};

export const reducer = (
  state: fromState.CandidateState = candidateInitialState,
  action: fromAction.CandidateActions
): fromUserState.CandidateState => {
  switch (action.type) {
    case fromAction.LOAD_CANDIDATES: {
      return {
        ...state,
        allcandidates: action.allCandidatesData
      };
    }
    case fromAction.LOAD_MATCHED_CANDIDATES: {
      return {
        ...state,
        matchedcandidates: action.matchedCandidatesData
      };
    }
    case fromAction.LOAD_ASSIGNED_CANDIDATES: {
      return {
        ...state,
        assignedcandidates: action.assignedCandidatesData
      };
    }

    default:
      return state;
  }
};
于 2018-05-24T07:29:10.793 回答