【发布时间】:2020-07-01 07:30:44
【问题描述】:
我有下一个错误“Functions are not valid as a React child React Native”,当我按下标题中的图像时,我试图显示一个抽屉(使用 react-navigation-stack 库创建),然后我使用react-navigation-drawer。到目前为止一切正常,当我尝试将变量导航传递给我的函数以在createStackNavigator 中使用时,问题就开始了,我不知道传递这个变量的正确方法是什么。
import React from 'react';
import {TouchableOpacity, Image} from "react-native";
import {createDrawerNavigator} from "react-navigation-drawer";
import {createStackNavigator} from 'react-navigation-stack';
//IMPORT SCENES
import HomeScreen from "../scenes/Home";
import UpdateProfileScreen from "../scenes/home/UpdateProfile";
const HomeStack = ({navigation}) => createStackNavigator(
{
Home: HomeScreen,
UpdateProfile: UpdateProfileScreen,
},
{
initialRouteName: 'Home',
defaultNavigationOptions: ({
headerLeft: () =>
<TouchableOpacity style={{alignItems: 'flex-end', margin: 16}}
onPress={() => navigation.openDrawer()}>
<Image style={{width: 24, height: 24}} source={require('../images/open-menu.png')}/>
</TouchableOpacity>
})
}
)
export default DrawerNavigator = createDrawerNavigator({
HomeStack
})
如何解决这个问题,我是 React Native 的新手。
【问题讨论】:
标签: react-native react-navigation react-navigation-stack react-navigation-drawer