导出默认功能应用程序({navigation})
有一个函数开头并包含
但是当我想使用导航功能并在onPress中使用“ navigation.navigate('Lead')
TypeError:未定义不是评估反应导航的对象
我得到一个错误。我的英文不是很好。请帮助我这个话题。现在,我向您添加示例代码。
import * as React from 'react';
import { StyleSheet, AsyncStorage, Alert, View, Image, TouchableOpacity } from 'react-native';
import { NavigationContainer } from '@react-navigation/native';
import { createStackNavigator } from '@react-navigation/stack';
import { createBottomTabNavigator } from '@react-navigation/bottom-tabs';
import { ApplicationProvider, Layout, Input, Button, Text, Spinner, IconRegistry, Icon } from '@ui-kitten/components';
import { EvaIconsPack } from '@ui-kitten/eva-icons';
import * as eva from '@eva-design/eva';
import LinkingConfiguration from './navigation/LinkingConfiguration';
import * as axios from 'axios';
import Base64 from './components/Base64';
import LeadScreen from './screens/LeadScreen';
import AddLeadScreen from './screens/AddLeadScreen';
import TabBarIcon from './components/TabBarIcon';
import HomeScreen from './screens/HomeScreen';
import CampaignsScreen from './screens/CampaignsScreen';
import MenuScreen from './screens/MenuScreen';
const Stack = createStackNavigator();
export default function App({ navigation }) {
const logout = async () => {
await AsyncStorage.removeItem('token');
dispatch({ type: 'SIGN_OUT' });
};
return (
<>
<IconRegistry icons={EvaIconsPack} />
<ApplicationProvider {...eva} theme={eva.light}>
<Layout style={styles.container}>
<AuthContext.Provider value={authContext}>
<NavigationContainer linking={LinkingConfiguration}>
<Stack.Navigator>
{state.isLoading ? (
// We haven't finished checking for the token yet
<Stack.Screen name="Splash" component={SplashScreen} options={{ headerShown: false }} />
) : state.token == null ? (
// No token found, user isn't signed in
<>
<Stack.Screen name="Login" component={LoginScreen} options={{ title: 'Oturum Açın' }} />
</>
) : (
// User is signed in
<>
<Stack.Screen name="User" component={UserScreen} options={{
headerStyle: { backgroundColor: '#e8a200' },
headerTintColor: 'white',
headerTitleStyle: { color: '#fff' },
headerRight: () => (
<TouchableOpacity activeOpacity={0.4} underlayColor='transparent' onPress={() => logout() } style={{ marginRight: 12 }}>
<Icon
style={styles.icon}
fill='#FFFFFF'
name='power-outline'
/>
</TouchableOpacity>
)
}} />
<Stack.Screen name="Lead" component={LeadScreen} options={{
title: 'Müşteri Adayları',
headerStyle: { backgroundColor: '#e8a200' },
headerTintColor: 'white',
headerTitleStyle: { fontWeight: 'bold', color: '#fff' },
headerRight: () => (
<TouchableOpacity activeOpacity={0.4} underlayColor='transparent' onPress={ () => console.log(navigation) } style={{ marginRight: 12 }}>
<Icon
style={styles.icon}
fill='#FFFFFF'
name='plus-outline'
/>
</TouchableOpacity>
)}} />
<Stack.Screen name="AddLead" component={AddLeadScreen} options={{ title: 'Müşteri Adayı Ekle' }} />
</>
)}
</Stack.Navigator>
</NavigationContainer>
</AuthContext.Provider>
</Layout>
</ApplicationProvider>
</>
);
}
答案 0 :(得分:1)
您可以像这样将导航或溃败道具传递到堆栈中的选项中。
options={({navigation})=>({
'Your code here'
})}
我已使用您的代码对其进行了编辑。
<Stack.Screen name="Lead" component={LeadScreen}
options={({ navigation }) => ({
title: 'Müşteri Adayları',
headerStyle: { backgroundColor: '#e8a200' },
headerTintColor: 'white',
headerTitleStyle: { fontWeight: 'bold', color: '#fff' },
headerRight: () => (
<TouchableOpacity activeOpacity={0.4}
underlayColor='transparent' onPress={() =>
console.log(navigation)} style={{ marginRight: 12 }}>
<Icon
style={styles.icon}
fill='#FFFFFF'
name='plus-outline'
/>
</TouchableOpacity>
)
})} />
答案 1 :(得分:0)
首先,我认为这里的navigation
是不必要的,您可以将其删除。
export default function App({ navigation }) {
在您的屏幕组件中,即UserScreen
,您可以像这样使用导航
function UserScreen({navigation}) {
// ...some codes
navigation.navigate('WHERE_TO_GO');
// ...some other codes
}
您还可以在不将导航传递给道具的情况下使用导航。
首先,您需要导入useNavigation
import { useNavigation } from '@'react-navigation/native'
然后,在您的组件中
function UserScreen() {
const nav = useNavigation();
// ...some codes
nav.navigate('WHERE_TO_GO');
// ...some other codes
}
您可以在这里https://reactnavigation.org/docs/use-navigation/
获得更多详细信息对于Lead
屏幕的onPress功能,
<Stack.Screen name="Lead" component={LeadScreen} options={({ navigation }) => ({
title: 'Müşteri Adayları',
headerStyle: { backgroundColor: '#e8a200' },
headerTintColor: 'white',
headerTitleStyle: { fontWeight: 'bold', color: '#fff' },
headerRight: () => (
<TouchableOpacity activeOpacity={0.4} underlayColor='transparent' onPress={ () => navigation.navigate('WHERE_TO_GO') } style={{ marginRight: 12 }}>
<Icon
style={styles.icon}
fill='#FFFFFF'
name='plus-outline'
/>
</TouchableOpacity>
)})} />
答案 2 :(得分:0)
您应该使用useLinkProps
来解决此问题:
import { useLinkProps } from '@react-navigation/stack';
这是有关如何编写使用该代码的示例:
const LinkButton = ({ to, action, children, ...rest }) => {
const { onPress, ...props } = useLinkProps({ to, action });
答案 3 :(得分:0)
我已经尝试了所有方法,但这对我有用。 在你的 app.js 中创建一个函数并传递 props 然后使用 props.navigation 访问,如下所示。
function CustomDrawerContent**(props)** {
.....
logOut = () => {
AsyncStorage.clear();
**props.navigation.navigate("Login");**
};
.....
}