How to navigate to other screen using the headerRight on the navigator code? - react-native

Question

The title is very confusing, but the explanation that I will give will be more clear.
I am creating one StackNavigator for my app, and I am defining one icon to be displayed on the header of one of my screens like so:
const navigator= createStackNavigator(
{
Initial: {
screen: Posts,
navigationOptions: {
title: "All Posts",
headerRight: () => {
return (
<TouchableOpacity style={{ padding: 5 }}>
<Feather name={"plus"} size={30} />
</TouchableOpacity>
);
},
},
},
Details: Details,
Create: Create,
},
{
initialRouteName: "Initial",
}
);
const App = createAppContainer(navigator);
export default () => {
return (
<Provider>
<App />
</Provider>
);
};
The problem is that I want to navigate the user to the Create screen when the user presses the icon that I am rendering on the right side of the header, but I do not know how to have access to the navigation.navigate() function that the navigator generates for all the screens. I know that I can define the header on the Posts' screen file, so I have access to the navigation.navigate() function, but I want to know if there is some way to do it on the App.js file.
EDIT
Reading the documentation I saw that the way that I am was creating the navigator is not what the official documentation recommends. I learned to make it like this by one old course, using React Navigation 4.x, and now with React Navigation 6.x I perceived the difference in the creation and changed the way that I was doing it on my app. You can check the documentation for the problem that I was having here

Answer 1

You can prepare your navigation option this way by sending props
options={(props) => ({
headerRight: () => <TouchableOpacity
onPress={()=>props.navigation.navigate('Screen')} style={{ padding: 5 }}>
<Feather name={"plus"} size={30} />
</TouchableOpacity>
})}