导出功能组件无效

时间:2019-12-01 14:45:35

标签: reactjs react-native functional-programming

我将名为Bus的屏幕作为函数而不是React类组件进行导出,并且始终会收到一个错误屏幕,上面写着:'Invariant Violation:元素类型无效:预期为字符串(对于内置组件)或类/ function(对于复合组件),但得到:未定义。您可能忘记了从定义文件中导出组件,或者可能混淆了默认导入和命名导入。检查Bus的渲染方法。'

对我来说毫无意义,因为我还在其他地方使用功能组件,而且效果很好!

我正在使用RN 0.60.5

我的公交页面,该错误提示我。

import React, { View, Text } from 'react';

import Maps from '../../components/Maps'
import SearchBox from '../../components/Searchbox'
import RouteDetailsBox from '../../components/RouteDetailsBox'

const Bus = () => (
    <View style={{ flex: 1 }}>
        <Maps />
        <SearchBox />
        <RouteDetailsBox />
    </View>
);

export default Bus;

App.js

/**
 * Sample React Native App
 * https://github.com/facebook/react-native
 *
 * @format
 * @flow
 */

import React from 'react';
import { Provider } from 'react-redux';
import store from './src/store';


//screens
import Bus from './src/screens/bus'

const App = () => {
  return (
    <Provider store={store}>
      <Bus />
    </Provider>
  );
};

export default App;

Error screen

1 个答案:

答案 0 :(得分:2)

来自

import React, { View, Text } from 'react';

import React from 'react';
import { View, Text} from 'react-native';