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如何调度多个动作创建者(React+Redux+服务器端渲染)

server-side-rendering redux reactjs javascript

grazdev · 技术社区 · 6 年前

我一直在学习一门关于如何使用React和Redux构建服务器端渲染应用程序的很好的课程,但我现在遇到的情况是,这门课程没有涵盖,我自己也无法理解。

请考虑以下组件(除了底部的导出部分外,它非常基本):

class HomePage extends React.Component {

    componentDidMount() {       
        this.props.fetchHomePageData();
    }   

    handleLoadMoreClick() {
        this.props.fetchNextHomePagePosts();
    }   

    render() {

        const posts = this.props.posts.homepagePosts; 
        const featuredProject = this.props.posts.featuredProject; 
        const featuredNews = this.props.posts.featuredNews; 
        const banner = this.props.posts.banner; 
        const data = ( posts && featuredProject && featuredNews && banner ); 

        if( data == undefined ) {
            return <Loading />; 
        }

        return(
            <div>
                <FeaturedProject featuredProject={ featuredProject } />
                <FeaturedNews featuredNews={ featuredNews } />
                <Banner banner={ banner } />                
                <PostsList posts={ posts } heading="Recently on FotoRoom" hasSelect={ true } />
                <LoadMoreBtn onClick={ this.handleLoadMoreClick.bind( this ) } />               
            </div>
        ); 

    }

}

function mapStateToProps( { posts } ) {
    return { posts }
}

export default {
    component: connect( mapStateToProps, { fetchHomePageData, fetchNextHomePagePosts } )( HomePage ),
    loadData: ( { dispatch } ) => dispatch( fetchHomePageData() )
};

以上操作很好:loadData函数发出API请求以获取一些数据,这些数据通过mapStateToProps函数馈送到组件中。但如果我想在同一个loadData函数中解雇多个动作创建者,该怎么办?唯一可行的是,如果我这样写函数:

function loadData( store ) {
    store.dispatch( fetchFeaturedNews() );
    return store.dispatch( fetchHomePageData() );
}

export default {
    component: connect( mapStateToProps, { fetchHomePageData, fetchNextHomePagePosts } )( HomePage ),
    loadData: loadData
};

但这并不好,因为我需要返回所有数据。。。请记住,导出的组件最终将处于以下路由配置中:

const Routes = [
    {
        ...App, 
        routes: [
            {
                ...HomePage, // Here it is!
                path: '/', 
                exact: true
            },
            {
                ...LoginPage, 
                path: '/login'
            },              
            {
                ...SinglePostPage, 
                path: '/:slug'
            },
            {
                ...ArchivePage, 
                path: '/tag/:tag'
            },                                      
        ]
    }
];

以下是在某条路线需要组件时如何使用loadData函数:

app.get( '*', ( req, res ) => {

    const store = createStore( req ); 

    const fetchedAuthCookie = req.universalCookies.get( authCookie ); 

    const promises = matchRoutes( Routes, req.path ).map( ( { route } ) => {
        return route.loadData ? route.loadData( store, req.path, fetchedAuthCookie ) : null;
    }).map( promise => {
        if( promise ) {
            return new Promise( ( resolve, reject ) => {
                promise.then( resolve ).catch( resolve ); 
            }); 
        }
    });

    ...

}

此外,这里还有一个动作创建者触发的动作示例。他们都承诺:

export const fetchHomePageData = () => async ( dispatch, getState, api ) => {

    const posts = await api.get( allPostsEP );

    dispatch({
        type: 'FETCH_POSTS_LIST', 
        payload: posts
    });             

}

和减速器:

export default ( state = {}, action ) => {
    switch( action.type ) {
        case 'FETCH_POSTS_LIST':
            return {
                ...state, 
                homepagePosts: action.payload.data 
            }                                       
        default: 
            return state; 
    }
}

1 回复 | 直到 6 年前

Tomasz Mularczyk 6 年前

因此,你的行动回报了一个承诺,你在问,你如何才能回报不止一个承诺。使用 Promise.all :

function loadData({ dispatch }) {
  return Promise.all([
    dispatch( fetchFeaturedNews() ),
    dispatch( fetchHomePageData() ),
  ]);
}

但是记住这个承诺。当所有的承诺都得到解决时,一切都会得到解决,它将返回一个大堆价值观:

function loadData({ dispatch }) {
  return Promise.all([
    dispatch( fetchFeaturedNews() ),
    dispatch( fetchHomePageData() ),
  ]).then(listOfResults => {
    console.log(Array.isArray(listOfResults)); // "true"
    console.log(listOfResults.length); // 2
  });
}

因此,您可能希望以不同的方式处理它。