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我在 react-navigation 标题中添加了一个抽屉菜单按钮来打开抽屉菜单。

该图标正常显示，但按下时返回引用的错误。

import React from 'react';
import { View, Dimensions } from 'react-native';
import { Button, Icon } from 'native-base';
import { createAppContainer, createStackNavigator, createDrawerNavigator } from 'react-navigation';
...

const DrawerConfig = {
  drawerWidth: Dimensions.get('window').width * 0.75,
  contentComponent: ({ navigation }) => {
    return(<MenuDrawer navigation={navigation} />)
  }
}

const HomeNavigator = createStackNavigator ({
...
}, {
  defaultNavigationOptions: ({ navigation }) => {
    return {
      headerTitleStyle: {
        fontWeight: 'bold'
      },
      headerLeft: (
        <Button transparent onPress={() => this.props.navigation.toggleDrawer()}>
          <Icon name='menu' style={{color: '#FFF'}} />
        </Button>
      ),
      headerRight: (
        <HomeIcon navigation={navigation} />
      ),
      headerStyle: {
        backgroundColor: '#b80003'
      },
      headerTintColor: '#FFF'
    }
  }
});

const DrawerNavigator = createDrawerNavigator (
  {
    'Principal': {
      screen: HomeNavigator
    },
    'Sobre o Aplicativo': {
      screen: InformationApp
    },
    'Sobre os Responsáveis': {
      screen: Team
    },
    'Sobre o Projeto': {
      screen: Project
    },
    'Política e Termos': {
      screen: Policy
    }
  },
  DrawerConfig
);

const AppDrawerContainer = createAppContainer(DrawerNavigator);

export default AppDrawerContainer;

需要注意的是，对于属于const DrawerNavigator = createDrawerNavigator我的屏幕，我使用上面相同的代码来呈现抽屉菜单按钮并且它工作正常，错误只发生在const screens HomeNavigator = createStackNavigator.

4

1 回答 1

-1

像这样更改您的代码

onPress={() => navigation.toggleDrawer()}>
于 2019-09-01T00:37:26.740 回答