disableOpenGesture隐藏不起作用导航抽屉反应原生

Question

我们在应用程序中使用导航抽屉来显示侧面菜单。在一些屏幕中，一旦用户尝试做左/右手势，我们不希望显示此导航抽屉。

所以，为此，我们试图隐藏一个特定的屏幕 - 手势/导航抽屉 - 但它不起作用。当用户滑动（左/右）时，抽屉仍然打开。

    const AppNavigator = StackNavigator(
      {
        // Drawer: { screen: Drawer },
        Register: {
          screen: Register,
          navigationOptions: ({ navigation }) => ({
            drawerLockMode: "locked-closed",
          })
        },
        TabHome: { screen: TabHome },
        Album: { screen: Album },
        offlineContent: { screen: offlineContent },
        changePassword: { screen: changePassword },
        Player: {screen: Player},
      },
      {
        initialRouteName: "TabHome",
        // header: null
      }
    );

我们尝试了drawerLockMode:"locked-closed"和disableOpenGesture: true，但没有任何工作。

注意：我使用的是"react-navigation": "^3.0.9"版本

任何禁止通过手势打开导航抽屉的建议？

Answer 1

你能试试这段代码吗？

const AppNavigator = StackNavigator(
      {
        // Drawer: { screen: Drawer },
        Register: {
          screen: Register
        },
        TabHome: { screen: TabHome },
        Album: { screen: Album },
        offlineContent: { screen: offlineContent },
        changePassword: { screen: changePassword },
        Player: {screen: Player},
      },
      {
        initialRouteName: "TabHome",
        // header: null
      }
    );
AppNavigator.navigationOptions = ({ navigation }) => {
  const navigationOptions = {};
  if (getCurrentRouteName(navigation.state) === 'Register') {
    navigationOptions.drawerLockMode = 'locked-closed';
  }
  return navigationOptions;
};

或者如果您的反应导航版本是V2，

const Stack = createStackNavigator({
    Register: Register,
},
    {
        drawerLockMode: 'locked-closed'
    }
);

getCurrentRouteName函数

getCurrentRouteName(navState) {

    if (navState.hasOwnProperty('index')) {
        this.getCurrentRouteName(navState.routes[navState.index])
    } else {
        console.log("Current Route Name:", navState.routeName)
        // can then save this to the state (I used redux)
        store.dispatch(setCurrentRouteName(navState.routeName))
    }

}