第二次操作调用时React / Redux State为空

时间:2016-04-21 08:07:01

标签: javascript node.js reactjs redux react-redux

对于React / Redux我还是比较新的,如果这是一个简单的问题我很抱歉,但我还没有找到解决方案。

我有两个动作:

// DESTINATIONS
// ==============================================

export const DESTINATION_REQUEST = 'DESTINATION_REQUEST';
export const DESTINATION_SUCCESS = 'DESTINATION_SUCCESS';
export const DESTINATION_FAILURE = 'DESTINATION_FAILURE';

export function loadDestination (params, query) {

    const state = params.state ? `/${params.state}` : '';
    const region = params.region ? `/${params.region}` : '';
    const area = params.area ? `/${params.area}` : '';

    return (dispatch) => {
        return api('location', {url: `/accommodation${state}${region}${area}`}).then((response) => {
            const destination = formatDestinationData(response);

            dispatch({
                type: DESTINATION_SUCCESS,
                destination
            });
        });
    };
}





// PROPERTIES
// ==============================================

export const PROPERTIES_REQUEST = 'PROPERTIES_REQUEST';
export const PROPERTIES_SUCCESS = 'PROPERTIES_SUCCESS';
export const PROPERTIES_FAILURE = 'PROPERTIES_FAILURE';

export function loadProperties (params, query, rows = 24) {

    return (dispatch, getState) => {

        console.log(getState());

        return api('search', {locationId: xxxxxx, rows: rows}).then((response) => {
            const properties = response.results.map(formatPropertiesData);

            dispatch({
                type: PROPERTIES_SUCCESS,
                properties
            });
        });
    };
}

这些与他们的相对减速器相结合:

// DESTINATIONS REDUCERS
// ==============================================

export default function destination (state = [], action) {
    switch (action.type) {
    case DESTINATION_SUCCESS:
        return action.destination;
    default:
        return state;
    }
}





// PROPERTIES REDUCERS
// ==============================================

export default function properties (state = [], action) {
    switch (action.type) {
    case PROPERTIES_SUCCESS:
        return action.properties;
    default:
        return state;
    }
}

从组件中调用它们(connectDataFetchers循环遍历被调用的操作并将它们返回到用于服务器端呈现的组件):



// PROPTYPES
// ==============================================

Search.propTypes = {
    destination: PropTypes.object.isRequired,
    properties: PropTypes.array.isRequired
};





// ACTIONS
// ==============================================

function mapStateToProps ({destination, properties}) {
    return {destination, properties};
}





// CONNECT & EXPORT
// ==============================================

export default connect(mapStateToProps)(
    connectDataFetchers(Search, [loadDestination, loadProperties])
);






export default function connectDataFetchers (Component, actionCreators) {
    return class DataFetchersWrapper extends React.Component {
        static propTypes = {
            dispatch: React.PropTypes.func.isRequired,
            location: React.PropTypes.object.isRequired,
            params: React.PropTypes.object.isRequired
        };
        
        static fetchData (dispatch, params = {}, query = {}) {
            return Promise.all(
                actionCreators.map((actionCreator) => dispatch(actionCreator(params, query)))
            );
        }

        componentDidMount () {
            DataFetchersWrapper.fetchData(
                this.props.dispatch,
                this.props.params,
                this.props.location.query
            );
        }

        render () {
            return (
                <Component {...this.props} />
            );
        }
    };
}
&#13;
&#13;
&#13;

我需要运行第一个操作(loadDestination),它将返回一个ID,然后需要传递给第二个操作以加载具有该位置ID的属性。

如果我对locationID进行硬编码,则此工作正常,但如果我尝试通过loadProperties访问getState()中的状态,则会返回{ destination: [], properties: [] }

有没有办法通过状态访问第一个动作的值?

按照@ pierrepinard_2

的建议进行管理以使其工作

我创建了一个新动作,按照我需要的顺序调度其他两个动作:

&#13;
&#13;
// SEARCH
// ==============================================

export function loadSearch (params, query) {
    
    return (dispatch) => {
        return dispatch(
            loadDestination(params, query)
        ).then(() => {
            return dispatch(
                loadProperties(params, query)
            )
        })
    }
}

// DESTINATIONS
// ==============================================

export const DESTINATION_REQUEST = 'DESTINATION_REQUEST';
export const DESTINATION_SUCCESS = 'DESTINATION_SUCCESS';
export const DESTINATION_FAILURE = 'DESTINATION_FAILURE';

export function loadDestination (params, query) {

    const state = params.state ? `/${params.state}` : '';
    const region = params.region ? `/${params.region}` : '';
    const area = params.area ? `/${params.area}` : '';

    return (dispatch) => {
        return api('location', {url: `/accommodation${state}${region}${area}`}).then((response) => {
            const destination = formatDestinationData(response);
            
            dispatch({
                type: DESTINATION_SUCCESS,
                destination
            });
        });
    };
}
  
  // PROPERTIES
// ==============================================

export const PROPERTIES_REQUEST = 'PROPERTIES_REQUEST';
export const PROPERTIES_SUCCESS = 'PROPERTIES_SUCCESS';
export const PROPERTIES_FAILURE = 'PROPERTIES_FAILURE';

export function loadProperties (params, query, rows = 24) {

    return (dispatch, getState) => {
        return api('search', {locationId: getState().destination.id, rows: rows}).then((response) => {
            const properties = response.results.map(formatPropertiesData);

            dispatch({
                type: PROPERTIES_SUCCESS,
                properties
            });
        });
    };
}
&#13;
&#13;
&#13;

然后在组件中我只请求一个动作:

&#13;
&#13;
export default connect(mapStateToProps)(
    connectDataFetchers(Search, [loadSearch])
);
&#13;
&#13;
&#13;

2 个答案:

答案 0 :(得分:3)

您在fetchData()方法中使用Promise.all():您的操作是并行调度的,而不是一个接一个地调度。

要确保您调用第一个目标,然后调用属性,您应该为搜索组件创建一个特定的异步操作创建器。在这种情况下,此异步操作创建者将实现您需要的连续请求。

答案 1 :(得分:0)

我同意@ pierrepinard_2。

使用bluebird的promise.map,您应该能够同步调用所有给定的承诺。

这个post on stack overflow可以帮助你解决这个问题。

让我们知道它是否有效

相关问题