如何在抽屉式导航器中未定义的屏幕中打开抽屉式导航器

Question

是否可以从我不想放入抽屉式导航仪的屏幕上打开抽屉式导航仪

const AppDrawerNavigator = createDrawerNavigator({
  Home:{
    screen:MainStackNavigator,//stack with tab
  },
  Pdf:{
    screen:AnotherScreen
  }
},
{
  drawerPosition: 'left',
  drawerWidth:Dimensions.get('window').width-50,

})

这是我的堆栈导航器

const AppStackNavigator = createStackNavigator({
  Main:{screen:AppDrawerNavigator,navigationOptions: {
    header: null,
  }},
  Login:{screen:LoginScreen},

  Map:{screen:MapScreen,navigationOptions:({navigation})=>({
    headerLeft:null,
    headerTransparent:true,
    headerRight:
        <TouchableOpacity
         onPress={()=>navigation.toggleDrawer()}>
           <IconMenu name='menu' size={25} style={{marginRight:20}} />
         </TouchableOpacity>})},
})

[我想让MapScreen打开我在上面定义的抽屉（它不起作用，它说导航未定义。当我单击IconMenu时，toggleDrawer（）不是函数）]

想你

Answer 1

使用它来打开或关闭抽屉

this.props.navigation.dispatch(DrawerActions.toggleDrawer());

在您的代码中，您可以用作
navigation.dispatch(DrawerActions.toggleDrawer());

Answer 2

您需要调度带有动作的NavigateAction来切换抽屉。

尝试一下：

  headerRight: () => {
    const navigateAction = NavigationActions.navigate({
      routeName: "Main",
      action: DrawerActions.toggleDrawer()
    });

    return (
        <TouchableOpacity
         onPress={()=> navigation.dispatch(navigateAction)}>
       <IconMenu name='menu' size={25} style={{marginRight:20}} />
     </TouchableOpacity>})},
    );
  }

并且您应该这样导入NavigateActions和DrawerActions：
import { createAppContainer, NavigationActions } from "react-navigation";
import { createDrawerNavigator, DrawerActions } from "react-navigation-drawer";

操作在routeName中执行，在您的情况下为DrawerNavigation
Here is the link to docs
And here is a snack example