Question

在我的Angular2应用程序中，我使用ngrx来管理状态，所以当我从服务器接收数据时，我会将一个动作发送给reducer。

MyExampleReducer.ts：

export const Reviews: ActionReducer<any> = (state: any[] = [], action: Action) => {
switch (action.type) {
        case GET_REVIEWS:
            return action.payload;
        case ADD_REVIEW :
            return [...state, {review : action.payload, replays : []}];
        case UPDATE_REVIEW:
            return  '' // return what ? 
        case DELETE_REVIEW:
            return state.filter(item => {
                return item.id !== action.payload.id;
            });
        default:
            return state;
    }
};

问题是当我必须更新我的评论数组中的项目时，以redux方式执行的最佳方法是什么？

Answer 1

假设您的州只是一个充满评论的数组，您可以执行以下操作：

export const Reviews: ActionReducer<any> = (state: any[] = [], action: Action) => {
switch (action.type) {
        case GET_REVIEWS:
            return action.payload;
        case ADD_REVIEW :
            return [...state, {review : action.payload, replays : []}];
        case UPDATE_REVIEW:

            // get an array of all ids and find the index of the required review
            let index = state.map(review => review.id)
                        .indexOf(action.payload.id);

            return [
              ...state.slice(0, index),
              Object.assign({}, state[index], action.payload),
              ...state.slice(index + 1)
            ]
        case DELETE_REVIEW:
            return state.filter(item => {
        return item.id !== action.payload.id;
        });
        default:
            return state;
}

首先，您需要找到应更新的评论索引。之后，您可以创建一个新数组，在该数组中替换索引位置的对象。

这种突变的一个很好的资源是video。

Answer 2

您可以使用map返回一个数组，该数组的元素与更新的操作相对应：

export const Reviews: ActionReducer<any> = (state: any[] = [], action: Action) => {
  switch (action.type) {
    case ADD_REVIEW:
      return [...state, { review: action.payload, replays: [] }];
    case UPDATE_REVIEW:
      return state.map(item => item.id === action.payload.id ? { review: action.payload, replays: [] } : item);
    case DELETE_REVIEW:
      return state.filter(item => item.id !== action.payload.id);
    default:
      return state;
    }
}

此外，您可以使用评论缩减程序来执行ADD_REVIEW和UPDATE_REVIEW操作来简化评论减少程序 - 评论缩减程序仅涉及管理评论列表而不是评论本身：

import { reviewReducer } from '...';
export const Reviews: ActionReducer<any> = (state: any[] = [], action: Action) => {
  switch (action.type) {
    case ADD_REVIEW:
      return [...state, reviewReducer(undefined, action)];
    case UPDATE_REVIEW:
      return state.map(item => item.id === action.payload.id ? reviewReducer(item, action) : item);
    case DELETE_REVIEW:
      return state.filter(item => item.id !== action.payload.id);
    default:
      return state;
    }
}

Angular 2 ngrx：在reducers中不可变地更新对象数组的正确方法是什么？

2 个答案: