Question

我有以下反应本机代码：

import React from 'react';
import {AppRegistry} from 'react-native';
import { createAppContainer } from 'react-navigation';
import { createStackNavigator} from 'react-navigation-stack';
import { createDrawerNavigator} from 'react-navigation-drawer';
import {View,Text,Button} from 'react-native';

class Sub extends React.Component {
  static navigationOptions = {
    headerTitle: () => (
      <Button
        onPress={() => this.props.navigation.dangerouslyGetParent().openDrawer()}
        title="Open Drawer"
      />
    ),
  };
  render() {
      const props = this.props;
      return (
        <View>
        <Text>{props.navigation.state.routeName}</Text>
        <Button onPress={_=>props.navigation.navigate("Sub1")} title="Sub1" />
        <Button onPress={_=>props.navigation.navigate("Sub2")} title="Sub2" />
        </View>
      );
  }
}

const SubRoutes = {
    "Sub1":Sub
};

const SubNavigator = createStackNavigator(SubRoutes);

const SubApp = createAppContainer(SubNavigator);

const MainRoutes = {
    "Main1":SubApp
};

const MainNavigator = createDrawerNavigator(MainRoutes);

const App = createAppContainer(MainNavigator);

AppRegistry.registerComponent("nav", () => App);

它将StackNavigator呈现为DrawerNavigator的子代。在Sub屏幕上，我有一个带有按钮的headerTitle，当单击时，我想打开抽屉式导航器的抽屉菜单。现在，单击按钮只会显示错误Sub.props.navigation不存在。

当单击标题标题按钮时，如何使抽屉菜单打开？

Answer 1

请为此修改您的navigationOptions代码：

  ...
  static navigationOptions = ({navigation}) => ({
    headerTitle: () => (
      <Button
        onPress={() => navigation.openDrawer()}
        title="Open Drawer"
      />
    )
  });
  ...

子导航器屏幕上的openDrawer（）

1 个答案: