I am not able to navigate to the other screen in my react native project the error is: navigation not found - react-native

Question

const Item = ({ item }) => (
<TouchableOpacity
onPress={() => this.navigation.navigate(
"carProfile",
{item: item}
)}>
<View style={styles.item}>
<Text style={styles.title}>{item.name}</Text>
<Text style={styles.details}>{item.details}</Text>
</View>
</TouchableOpacity>
);
const List = (props,navigation) => {
const renderItem = ({ item }) => {
if (props.searchPhrase === "") {
return <Item item={item} />;
}
if (item.name.toUpperCase().includes(props.searchPhrase.toUpperCase().trim().replace(/\s/g, ""))) {
return <Item item={item} />;
}
if (item.details.toUpperCase().includes(props.searchPhrase.toUpperCase().trim().replace(/\s/g, ""))) {
return <Item item={item} />;
}
};
return (
<SafeAreaView style={styles.list__container}>
<View
onStartShouldSetResponder={() => {
props.setClicked(false);
}}
>
<FlatList
data={props.data}
renderItem={renderItem}
keyExtractor={(item) => item.id}
/>
</View>
</SafeAreaView>
);
};
export default List;
the error is continuing to be there despite me trying to put navigation container in different positions. I want to send the user to the carProfile page, where the data passed in item is reused. This way user can know about the selection they are looking for

Answer 1

Use this way
onPress={() => this.props.navigation.navigate(
"carProfile",
{item: item}
)}

Answer 2

The navigation is inside the props, not as a second argument. Below is a classic example. And also note, you cannot extract navigation prop for child components, it will only work on Main components(Screens).
function HomeScreen({ navigation }) {
return (
<View style={{ flex: 1, alignItems: 'center', justifyContent: 'center' }}>
<Text>Home Screen</Text>
<Button
title="Go to Details"
onPress={() => navigation.navigate('Details')}
/>
</View>
);
}
If you want to use navigation inside child components, you can use the navigation hooks.
import { useNavigation } from "#react-navigation/native";
function MyBackButton() {
const navigation = useNavigation();
return (
<Button
title="Back"
onPress={() => {
navigation.goBack();
}}
/>
);
}