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 /> }
]

我刚刚花了30分钟来解决这个基本的问题。

我的问题是我导入react原生元素

例如导入React，{文本，图像，组件}从' React ';

并试图使用它们，这导致我收到这个错误。

一旦我从<文本>切换到<p>和<图像>到<img>，一切都按预期工作。

EDIT

你把过程复杂化了。只要这样做:

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';

ReactDom.render(<routes />, document.getElementById('root'));

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;

在我的例子中，在使用Jest时，我模拟了一些模块，它导致了错误，因为模拟的模块不包括在其他领域使用的某些组件。

因此，请检查您的测试组件中是否有一些模拟函数。

// @flow

import React from 'react';
import { styleLocal } from './styles';
import {
  View,
  Text,
  TextInput,
  Image,
} from 'react-native';
import { TouchableOpacity } from 'react-native-gesture-handler';

export default React.forwardRef((props, ref) => {
const { onSeachClick, onChangeTextSearch , ...otherProps } = props;

  return (
        <View style={styleLocal.sectionStyle}>
        <TouchableOpacity onPress={onSeachClick}>
            <Image                              
                    source={require('../../assets/imgs/search.png')}
                    style={styleLocal.imageStyle} />
                </TouchableOpacity>
            <TextInput
                style={{ flex: 1, fontSize: 18 }}
                placeholder="Search Here"
                underlineColorAndroid="transparent"
                onChangeText={(text) => { onChangeTextSearch(text) }}
            />
        </View>
  );
});
 import IGPSSearch from '../../components/IGPSSearch';
<Search onSeachClick={onSeachClick} onChangeTextSearch= {onChangeTextSearch}> </Search>

所以我遇到了这个问题，我用一种奇怪的方式解决了它。我想我应该把它扔了，以防其他人遇到它，变得绝望。

我有一个有条件地呈现react元素，它只在对象状态大于零时才会呈现。在呈现函数中-当三元运算符被证明为真时，我收到了这篇文章所涉及的错误。我使用了一个<React。片段>在我的渲染函数-这是我通常分组子的方式，而不是更新的方式与更新的短语法(<>和</>)。

However, when I changed the <React.Fragment> to the short syntax - the react element rendered correctly and the error stopped appearing. I took a look at Facebooks React-Fragment documentation, and I don't see any reason on there to believe there is a dramatic difference between the two that would cause this. Regardless, below are two simplified snippets of code for your review. The first one demonstrates the code that was yielding the error, and the second one demonstrates the code that fixed the issue. At the bottom is the render function of the component itself.

希望有人觉得这有帮助。

代码示例1 -错误 有条件渲染的React元素

  renderTimeFields = () => {
    return (
      <React.Fragment>
        <div className="align-left">
                <div className="input-row">
                  <label>Date: </label>
                  <input type="date" className="date" name="date" min={new Date().toISOString().split('T')[0]} required/>
                </div>
              </div>
              <div className="align-left sendOn-row">
                <div className="input-row">
                  <label>Time: </label>
                  <input type="time" className="time" name="time" required/>
                </div>
              </div>
      </React.Fragment>
    );
  }

代码示例2 -无错误 有条件渲染的React元素

  renderTimeFields = () => {
    return (
      <>
        <div className="align-left">
                <div className="input-row">
                  <label>Date: </label>
                  <input type="date" className="date" name="date" min={new Date().toISOString().split('T')[0]} required/>
                </div>
              </div>
              <div className="align-left sendOn-row">
                <div className="input-row">
                  <label>Time: </label>
                  <input type="time" className="time" name="time" required/>
                </div>
              </div>
      </>
    );
  }

在组件中呈现函数

render() {
<form>
...
{emailsInActiveScript > 1 ? this.renderTimeFields() : null}
</form>
etc...
}