reactjs - React Router Lazy 组件不起作用

Question

所以这有效:

import Page from 'components/Page';
...

render() {
   return (
     <Route render={props => <Page {...props}/>}/>
   );
}

但这不是:

import React, { lazy } from 'react';
const Cmp = lazy(() => import('components/Page'));
...

render() {
   return (
     <Route render={props => <Cmp {...props}/>}/>
   );
}

react 16.8.6React 路由器 5.0.0

我明白了:

The above error occurred in one of your React components:
    in Unknown (at configured.route.js:41)
    in Route (at configured.route.js:41)
    in ConfiguredRoute (created by Context.Consumer)
    ...rest of stack trace

谁能看出我在这里做了什么蠢事吗？

Answer 1

引用 React Docs on code splitting ，建议使用Suspense具有定义的 fallback因此，当组件未加载时，您可以渲染一些内容来代替组件。

// Direct paste from https://reactjs.org/docs/code-splitting.html

import { BrowserRouter as Router, Route, Switch } from 'react-router-dom';
import React, { Suspense, lazy } from 'react';

const Home = lazy(() => import('./routes/Home'));
const About = lazy(() => import('./routes/About'));

const App = () => (
  <Router>
    <Suspense fallback={<div>Loading...</div>}>
      <Switch>
        <Route exact path="/" component={Home}/>
        <Route path="/about" component={About}/>
      </Switch>
    </Suspense>
  </Router>
);

Suspense 应该是 <Route> 的父级使用 lazy 的元素