尝试得到react-router (v4.0.0)和react-hot loader (3.0.0-beta.6)很好地发挥,但在浏览器控制台得到以下错误:

Warning: React.createElement: type is invalid -- expected a string
(for built-in components) or a class/function (for composite
components) but got: undefined. You likely forgot to export your
component from the file it's defined in.

index.js:

import React from 'react';
import ReactDom from 'react-dom';
import routes from './routes.js';
require('jquery');
import 'bootstrap/dist/css/bootstrap.min.css';
import 'bootstrap/dist/js/bootstrap.min.js';
import './css/main.css';

const renderApp = (appRoutes) => {
    ReactDom.render(appRoutes, document.getElementById('root'));
};

renderApp( routes() );

routes.js:

import React from 'react';
import { AppContainer } from 'react-hot-loader';
import { Router, Route, browserHistory, IndexRoute } from 'react-router';
import store from './store/store.js';
import { Provider } from 'react-redux';
import App from './containers/App.jsx';
import Products from './containers/shop/Products.jsx';
import Basket from './containers/shop/Basket.jsx';

const routes = () => (

    <AppContainer>
        <Provider store={store}>
            <Router history={browserHistory}>
                <Route path="/" component={App}>
                    <IndexRoute component={Products} />
                    <Route path="/basket" component={Basket} />
                </Route>
            </Router>
        </Provider>
    </AppContainer>

);

export default routes;

当前回答

您需要了解命名导出和默认导出。我什么时候应该用大括号导入ES6 ?

在我的例子中,我通过改变

import Provider from 'react-redux'

to

import { Provider } from 'react-redux'

其他回答

在我的例子中,我添加了相同的自定义组件作为道具,而不是实际的道具

   import AddAddress from '../../components/Manager/AddAddress';
    
    class Add extends React.PureComponent {
      render() {
        const {
          history,
          addressFormData,
          formErrors,
          addressChange,
          addAddress,
          isDefault,
          defaultChange
        } = this.props;
    

错误:

    return (
        <AddAddress
          addressFormData={addressFormData}
          formErrors={formErrors}
          addressChange={addressChange}
          addAddress={AddAddress} // here i have used the Component name as probs in same component itself instead of prob
          isDefault={isDefault}
          defaultChange={defaultChange}
        />
      </SubPage>

解决方案:

return (
    <AddAddress
      addressFormData={addressFormData}
      formErrors={formErrors}
      addressChange={addressChange}
      addAddress={addAddress} // removed the component name and give prob name
      isDefault={isDefault}
      defaultChange={defaultChange}
    />
  </SubPage>

我已经遇到过这个问题了。我的解决方案是:

在文件配置路由:

const routes = [
    { path: '/', title: '', component: Home },
    { path: '*', title: '', component: NotFound }
]

to:

const routes = [
    { path: '/', title: '', component: <Home /> },
    { path: '*', title: '', component: <NotFound /> }
]

在我的例子中,我的组件中有文本组件,当我像这样导入没有花括号的文本时,这就是错误的原因

import Text from '..';  <-wrong

为了修复这个错误,我像这样在花括号内导入Text

import {Text} from '..';  <-right

我的情况不像上面许多人说的那样是一个重要的问题。在我的版本中,我们使用了一个包装器组件来做一些转换逻辑,我错误地传递了子组件,就像这样:

const WrappedComponent = I18nWrapper(<ChildForm {...additionalProps} />);

当我应该把它作为一个函数传递的时候:

const WrappedComponent = I18nWrapper(() => <ChildForm {...additionalProps} />);

我所缺少的是我正在使用

import { Router, Route, browserHistory, IndexRoute } from 'react-router';

正确答案应该是:

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

当然你需要添加npm包react-router-dom:

npm install react-router-dom@next --save