TypeError:无法读取未定义的属性(读取‘props’)
2022-05-22
6005
我正在使用 React Native 制作社交网络应用程序。用户可以拍照,单击保存按钮以将图像 uri 作为 prop 传递到保存屏幕。这是代码 -
App.js
const store = createStore(rootReducer, applyMiddleware(thunk))
const Stack = createStackNavigator()
export default function App() {
const [loading, setLoading] = useState(false)
const [loggedIn, setLoggedIn] = useState(false)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
return (
<Provider store={store}>
<NavigationContainer>
<Stack.Navigator initialRouteName="HomeScreen">
<Stack.Screen
name="HomeScreen"
component={HomeScreen}
options={{headerShown: false}}
/>
<Stack.Screen
name="CreatePostScreen"
component={CreatePostScreen}
options={{title: "New Post!"}}
navigation={this.props.navigation} //error line
/>
<Stack.Screen
name="SaveImageScreen"
component={SaveImageScreen}
options={{title: ""}}
/>
</Stack.Navigator>
</NavigationContainer>
</Provider>
)
}
CreatePostScreen.js
function CreatePostScreen({navigation}) {
const [hasCameraPermission, setHasCameraPermission] = useState(null)
const [hasGalleryPermission, setHasGalleryPermission] = useState(null)
const [type, setType] = useState(Camera.Constants.Type.back)
const [camera, setCamera] = useState(null)
const [image, setImage] = useState(null)
useEffect(() => {
~~~~~~~~~~~~~~~
}, [])
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
const takePicture = async () => {
if (camera) {
const data = await camera.takePictureAsync(null)
setImage(data.uri)
}
}
return (
<View style={{flex: 1}}>
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`
<IconButton
icon="radiobox-marked"
color={Colors.SEA_BLUE}
size={35}
onPress={() => takePicture()}
/>
<IconButton
icon="check-circle"
color={Colors.SEA_BLUE}
size={35}
onPress={() => navigation.navigate("SaveImageScreen", {image})} //saved image uri is passed to another screen as prop
/>
</View>
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
</View>
)
}
CreatePostScreen 和 SaveImageScreen 都是 App.js 中导航容器下的堆栈屏幕。
但每当我运行应用程序时,我都会收到此错误 ~
TypeError: Cannot read properties of undefined (reading 'props')
或者,有时是这个错误 ~
TypeError: undefined is not an object (evaluating 'this.props')
我应该在代码中做哪些更改?
3个回答
为什么使用
this.props
?
这不是基于类的组件。
您正在使用函数式,您应该编写类似以下内容的内容:
export default function App(props) {{
并使用
props.navigation
owenizedd
2022-05-22
删除
navigation={this.props.navigation}
AE0011
2022-05-22
您需要删除此
navigation={this.props.navigation} //error line
此 关键字用于基于类的组件,并且您的组件是功能组件,并且您不需要传递导航(我知道您想要实现的目标)为了浏览组件,您只需使用来自 react-navigation 的钩子即可,如下所示
import * as React from 'react';
import { Button } from 'react-native';
import { useNavigation } from '@react-navigation/native';
function MyBackButton() {
const navigation = useNavigation();
return (
<Button
title="Back"
onPress={() => {
navigation.goBack();
}}
/>
);
}
Mandip Giri
2022-05-23