在没有屏幕的情况下使用抽屉导航

Question

我有一个关注stack navigator

<Stack.Navigator>
  <Stack.Screen name="Home" component={HomeScreen} />
  <Stack.Screen name="Settings" component={SettingsScreen} />
  <Stack.Screen name="AboutUs" component={AboutUsScreen} />
  <Stack.Screen name="ContactUs" component={ContactUsScreen} />
  <Stack.Screen name="Blog" component={BlogScreen} />
</Stack.Navigator>

我想要 drawer（带有自定义抽屉内容）而不使用 nesting drawer navigators。

Answer 1

我看到了你的问题，我正在努力帮助你。我为抽屉组件做了定制设计。 - 首先，您可以为抽屉设计创建一个额外的文件，例如 DrawerComponent.js 并在您创建抽屉导航器的代码中导入

import DrawerComponent from "./DrawerComponent";
const Primary_Nav = createDrawerNavigator(
{
  screen1: {
  screen: screen1,
  navigationOptions: {
  drawerLabel: () => null
   }
 },
screen2: {
  screen: screen2,
  navigationOptions: {
    drawerLabel: "detail"
  }
},
 {
 initialRouteName: "screen1",
drawerPosition: "left",
drawerType: "slide",
 contentComponent: DrawerComponent //<<< i added this
}
);

const PrimaryNav = createAppContainer(Primary_Nav);
export default PrimaryNav;

现在，在 DrawerComponent.js 中，您可以自定义组件内容。

import React, { Component } from "react";
import { Text, View, TouchableOpacity } from "react- 
native";

export default class DrawerComponent extends Component {
  constructor(props) {
    super(props);

    this.state = {

    };
  }

  render() {
    const { navigation } = this.props;
    return (
      <View style={{ flex: 1, paddingVertical: 40,paddingHorizontal: 20 }}>
        <TouchableOpacity
          style={{ margin: 20 }}
          onPress={() => navigation.navigate("screen1")}
        >
          <Text>Home</Text>
        </TouchableOpacity>
        <TouchableOpacity
          style={{ margin: 20 }}
          onPress={() => navigation.navigate("screen2")}
        >
          <Text>Detail</Text>
        </TouchableOpacity>
  </View>
  );
 }
}

希望对您有所帮助。谢谢你，祝你编码愉快。