How to display a button at the bottom of a Webview in react-native? - react-native

Question

Inside my component (PrivacyPolicy.js), i have a header view, a webview, and a footer view. the webview, depending on the size, gets scrollable. my issue is that the footer view is displayed at the bottom of the screen like if its style was "position: 'absolute'" so it keeps displayed while scrolling. I need to have it after all webview is displayed.
<View style={styles.main_container}>
<View style={styles.header_container}>
...
</View>
<WebView originWhitelist={['*']} source={{ html: privacyPolicyContent }}/>
<View style={styles.footer_container}>
<CheckBox
disabled={false}
value={this.state.isChecked}
onValueChange={(newValue) => this.setState({
isChecked: newValue
})}
style={styles.checkbox}
tintColors={{ true: '#157dfa' }}
/>
<Text style={styles.checkbox_text}>I have read and accept the Privacy Polic</Text>
</View>
</View>
My styles:
const styles = StyleSheet.create({
main_container: {
flex: 1,
paddingHorizontal:'5%'
},
header_container: {
height: scale(90),
flexDirection: 'row',
marginLeft: 10
},
checkbox_container: {
flexDirection: 'row'
},
checkbox: {
marginLeft: -5,
},
checkbox_text: {
marginTop: 8,
fontSize: 10
}
})

Answer 1

I can see few suggestions:
Since your button is a React Native Button => You can show/hide based on the scrollY positions. For that, you need to communicate over the Bridge to dispatch an event accordingly.
As an alternative solution => You can create the button on the Webview its self to have the same functionality.