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;
// @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>
组件数组
获得此错误的常见方法是使用组件数组，并使用位置索引从数组中选择要呈现的组件。我多次看到这样的代码:
const checkoutSteps = [Address, Shipment, Payment]
export const Checkout = ({step}) => {
const ToRender = checkoutSteps[step]
return (
<ToRender />
)
}
这不是必要的坏代码，但如果你用一个错误的索引(例如-1，或者在这种情况下是3)调用它，ToRender组件将是未定义的，抛出React。createElement:类型无效…错误:
<Checkout step={0} /> // <Address />
<Checkout step={1} /> // <Shipment />
<Checkout step={2} /> // <Payment />
<Checkout step={3} /> // undefined
<Checkout step={-1} /> // undefined
合理的解决方案
你应该使用更明确的方法保护自己和同事不受这些难以调试的代码的影响，避免使用神奇的数字并使用PropTypes:
const checkoutSteps = {
address: Address,
shipment Shipment,
payment: Payment
}
const propTypes = {
step: PropTypes.oneOf(['address', 'shipment', 'payment']),
}
/* TIP: easier to maintain
const propTypes = {
step: PropTypes.oneOf(Object.keys(checkoutSteps)),
}
*/
const Checkout = ({step}) => {
const ToRender = checkoutSteps[step]
return (
<ToRender />
)
}
Checkout.propTypes = propTypes
export default Checkout
你的代码看起来是这样的:
// OK
<Checkout step="address" /> // <Address />
<Checkout step="shipment" /> // <Shipment />
<Checkout step="payment" /> // <Payment />
// Errors
<Checkout step="wrongstep" /> // explicit error "step must be one of..."
<Checkout step={3} /> // explicit error (same as above)
<Checkout step={myWrongVar} /> // explicit error (same as above)
这种方法的好处
代码更明确，你可以清楚地看到你想呈现什么
你不需要记住数字和它们隐藏的含义(1代表地址，2代表…)
错误也是显式的
不会让你的同事头疼:)
