React Router v4 嵌套路由 props.children

Question

我正在更新我的通用 react redux 应用程序以使用 react router v4。我在主要布局路线下有嵌套路线。以前我使用 {props.children} 来显示子路由的内容，但这不再起作用了。这在 V4 中如何工作？

<Provider store={store} key="provider">
  <div>
    <Route component={Layout} />
    <Switch>
      <Route path="/" component={HomePage} />
      <Route component={Error404} />
    </Switch>
  </div>
</Provider>

或

<Provider store={store} key="provider">
  <Layout>
    <Route path="/" component={HomePage} />
    <Route component={Error404} />
  </Layout>
</Provider>

这就是我的布局文件的样子

const Layout = props => (
 <div className="o-container">
   <Header />
     <main>
      {props.children}
     </main>
   <Footer />
 </div>
);

Answer 1

只是觉得我必须分享这个。如果您在 Header 组件中使用 Link 组件。上面的答案是行不通的。您必须再次将BrowserRouter 设为父级以支持Link。变成这样：

<BrowserRouter>
 <Layout>
  <Switch>
    <Route exact path="/" component={HomePage} />
    <Route path="/other" component={OtherComponent} />
    <Route component={Error404} />
  </Switch>
 </Layout>
</BrowserRouter>

Answer 2

请仔细阅读此博客。 https://codeburst.io/react-router-v4-unofficial-migration-guide-5a370b8905a

没有了<IndexRoute>

在 v3 中允许路由到顶级路径上的某个组件的组件：

// in src/MyApp.js
const MyApp = () => (
    <Router history={history}>
        <Route path="/" component={Layout}>
            <IndexRoute component={Dashboard} />
            <Route path="/foo" component={Foo} />
            <Route path="/bar" component={Bar} />
        </Route>
    </Router>
)

这个组件在 v4 中不再存在。要替换它，请结合使用 、exact 和 route ordering（将索引路由放在最后）：

  // in src/MyApp.js
const MyApp = () => {
    <Router history={history}>
        <Route path="/" component={Layout} />
    </Router>
}
// in src/Layout.js
const Layout = () => (
    <div className="body">
        <h1 className="title">MyApp</h1>
        <div className="content">
            <Switch>
                <Route exact path="/foo" component={Foo} />
                <Route exact path="/bar" component={Bar} />
                <Route exact path="/" component={Dashboard} />
            </Switch>
        </div>
    </div>
);

Answer 3

我会使用这种结构，没有 props.children ：

const Main = () => (
  <main>
    <Switch>
      <Route exact path="/" component={HomePage} />
      <Route component={Error404} />
    </Switch>
  </main>
)

const Layout = () => (
  <div>
    <Header />
    <Main />
    <Footer />
  </div>
)

ReactDOM.render((
    <Provider store={store}>
      <BrowserRouter>
        <Layout />
      </BrowserRouter>
   </Provider>
), document.getElementById('root'))

Answer 4

如果与 Redux 一起使用，则不带 Switch 元素

AppRouter.js

import React from 'react'
import { BrowserRouter as Router, Route, Link } from 'react-router-dom'



const AppRouter = () => (
  <Layout>
    <Router>
      <div>
        <nav>
          <ul>
            <li><Link to="/">Home</Link></li>
            <li><Link to="/about">About</Link></li>
            <li><Link to="/contact">Contact</Link></li>
          </ul>
        </nav>
        <Route exact path="/" component={Home}/>
        <Route path="/about" component={About}/>
        <Route path="/contact" component={Contact}/>
      </div>
    </Router>
  </Layout>
)

export default AppRouter;

布局.js

const Layout = props => ({
  render() {
    return (
      <div className="container">
        <Header />
        <main>{props.children}</main>
        <Footer />
      </div>
    );
  }
});

export default Layout;

Provider 放置在 Render 函数中

import React from 'react';
import ReactDOM from 'react-dom';
import { Provider } from 'react-redux';
import { createStore, applyMiddleware } from 'redux';


import AppRouter from './AppRouter';
import reducers from './reducers';

const createStoreWithMiddleware = applyMiddleware()(createStore);

ReactDOM.render(
  <Provider store={createStoreWithMiddleware(reducers)}>
    <AppRouter />
  </Provider>
  , document.getElementById('app'));

Answer 5

添加到@Dez 答案

完整的原生/核心实现，支持 Redux

index.js

import React from 'react';
import ReactDOM from 'react-dom';
import { Provider } from 'react-redux';
import { Router, Route, Switch } from 'react-router';
import createMemoryHistory from 'history/createMemoryHistory';

const history = createMemoryHistory();

import App from './components/App';
import Home from './components/Home';
import Login from './components/Login';
import store from './store';

ReactDOM.render((
  <Provider store={ store }>
    <Router history={history}>
      <App>
        <Switch>
          <Route exact path="/" component={Home} />
          <Route path="/login" component={Login} />
        </Switch>
      </App>
    </Router>
  </Provider>
), document.getElementById('root'));

App.js

import Header from './Header';
import Home from './Home';
import React from 'react';
import {connect} from 'react-redux';

const mapStateToProps = state => ({appName: state.appName});

class App extends React.Component {
  render() {
    return (
      <div >
        <Header appName={this.props.appName} /> {/*common header*/}
        {this.props.children}
      </div>
    );
  }
}

export default connect(mapStateToProps, () => ({}))(App);

Answer 6

我采用了<Provider>out，因为它属于react-redux，并且您不需要它作为使用react-router 进行路由的基础（无论如何您可以轻松地添加它来封装结构）。

在 React Router V4 中，Router 已重命名为 BrowserRouter 并从包 react-router-dom 中导入。因此，对于嵌套路由，您需要将其作为 <Layout> 的子级插入。

index.js

import { Switch, Route } from 'react-router';
import { BrowserRouter } from 'react-router-dom';
import Layout from './Layout';
...
const Root = () => (
  <Layout>
    <BrowserRouter>
      <Switch>
        <Route exact path="/" component={HomePage} />
        <Route path="/other" component={OtherComponent} />
        <Route component={Error404} />
      </Switch>
    </BrowserRouter>
  </Layout>
);

ReactDOM.render(
  <Root/>,
  document.getElementById('root')
);

布局.js

import React from 'react';
import Header from './Header';
import Footer from './Footer';

const Layout = props => ({
  render() {
    return (
      <div className="o-container">
        <Header />
        <main>{props.children}</main>
        <Footer />
      </div>
    );
  }
});

export default Layout;

考虑一下，这仅适用于网络。本机实现不同。
我在 Create React App 中上传了 a small project，我在其中展示了 V4 中嵌套路由的实现。