How can we create a react-native component which can be displayed with .show() method - react-native

Question

I am looking for pointers where I can start from.
I want to create a react native feedback form which can be displayed with .show method.
for eg:
export class FeedbackComponent extends React.component{
show() {
// define this method in a way so that can be called from outside as FeedbackComponent.show()
// which eventually create a new screen with below rendered View
}
render (){
return <View>Feedback Form</View>
}
}
I should be able to use this Component in any other component as
import FeedbackComponent from './FeedbackComponent'
new FeedbackComponent.show()

Answer 1

I would always start with considering the application state. UI in React is updated whenever the State of our components changes.
In your case I would have to think of the parent context in which your feedback form will need to be displayed. In its simplest form, this context will likely be a parent screen-component within which your FeedbackForm component is either shown or hidden.
I've made a Snack of a simple implementation you can find it here: https://snack.expo.io/#stephos_/show-feedback-form
In my case, the App component is the parent screen-component within which we need to render or not render a FeedbackForm Component.
So I would start with adding the relevant state property to the App (parent screen) component like so:
state = {
feedbackFormVisible : false
}
I would then define a method within the same parent class in order to toggle the state when we need to like so:
handleFeedbackFormVisibility = () => this.setState(prevState => ({feedbackFormVisible:!prevState.feedbackFormVisible}))
This handler takes in the previous state in our parent component and toggles the value of the feedbackFormVisible property (i.e. from false to true).
In my case, I call this handler every time we press a Button component like so:
<Button title="Give Feedback" onPress={this.handleFeedbackFormVisibility}/>
You could however trigger the same handler and update the state of the parent component in any other way (i.e. after a timer expires or after a specific scroll point is passed).
The App Component's render method will then decide if the FeedbackForm component should be displayed based on the value of the feedbackFormVisible in our App Component's state. We achieve this by wrapping our FeedbackForm component within an Elvis Conditional within the render method which will return the the appropriate UI (i.e. either with a visible feedback form or not):
{ this.state.feedbackFormVisible ? () : null}
Below the full App component code:
import * as React from 'react';
import { Text, View, StyleSheet, Button } from 'react-native';
import FeedbackForm from './components/FeedbackForm';
export default class App extends React.Component {
state = {
feedbackFormVisible : false
}
handleFeedbackFormVisibility = () => this.setState(prevState => ({feedbackFormVisible:!prevState.feedbackFormVisible}))
render() {
return (
<View style={styles.container}>
<Text style={styles.paragraph}>
This is the App Parent Component
</Text>
{
this.state.feedbackFormVisible ?
(<FeedbackForm />)
: null
}
<Button title="Give Feedback" onPress={this.handleFeedbackFormVisibility}/>
</View>
);
}
}
And below the Feedback Component code (Notice that the shown/hidden logic is actually handled in the parent component not in here):
import * as React from 'react';
import { Text, View, StyleSheet } from 'react-native';
export default class FeedbackForm extends React.Component {
render() {
return (
<View style={styles.container}>
<Text style={styles.paragraph}>
This is the feedback form!
</Text>
</View>
);
}
}