在使用开关导航绑定导航时返回此错误。
我试过删除它。 然后 props 返回未定义的"导航"。
import React from 'react';
import { Text, View, Image, TouchableOpacity } from 'react-native';
import Icon from 'react-native-vector-icons/Ionicons';
import { createSwitchNavigator, createAppContainer , withNavigation } from 'react-navigation';
import {widthPercentageToDP as wp, heightPercentageToDP as hp} from 'react-native-responsive-screen';
import layout from '../../constants/LayoutStyle'
import QuoteScreen from './QuoteScreen';
const HomeScreen = () => {
return (
<View style={styles.viewStyles}>
<View style={{position: 'absolute',top: hp('50%')+wp('37.5%'),left:wp('15%'),width: wp('32.5%'),height:
wp('32.5%'),backgroundColor:'rgba(255,255,255,0.1)'}}>
<View style={{alignItems: 'center',justifyContent: 'center',flex:1}}>
<Icon name="ios-book" color="purple" size={wp('10%')}
onPress={() => this.props.navigation.navigate('Quote')}
/>
<Text style={styles.tabTextStyle}>Books</Text>
</View>
</View>
);
};
const RootStack = createSwitchNavigator(
{
Home: HomeScreen,
Quote: QuoteScreen,
}
);
const AppContainer = createAppContainer(RootStack);
export default class app extends React.Component {
render() {
return <AppContainer />
}
}
预计正确完成导航
HomeScreen 是一个功能组件,因此您不应该使用 this.props.navigation,只需说 props.navigation。
如果你想在函数组件中使用 props,那么 use 应该将 props 作为参数传递给该功能组件。像这样=>
const HomeScreen = (props) => {
return (
<View style={styles.viewStyles}>
<View style={{position: 'absolute',top:
hp('50%')+wp('37.5%'),left:wp('15%'),width: wp('32.5%'),height:
wp('32.5%'),backgroundColor:'rgba(255,255,255,0.1)'}}>
<View style={{alignItems: 'center',justifyContent: 'center',flex:1}}>
<Icon name="ios-book" color="purple" size={wp('10%')}
onPress={() => this.props.navigation.navigate('Quote')}
/>
<Text style={styles.tabTextStyle}>Books</Text>
</View>
</View>
);
};
如果这不起作用,则将导航作为道具传递给您使用的 HomeScreen 组件,就像这样 =>
<HomeScreen
navigation = {this.props.navigation} // or navigation = {props.navigation}, if it is functional component
/>