how to create action button react native redirect other page?

Question

i'm new in react native. i'm use react native action button and i want if button clicked, then show other page. this is my code but still doesn't work. have any solution?

render() {
    return (
        <View style={styles.container}>
           <ActionButton buttonColor="#1E73C1" onPress={() => this.buttonPressed}>
          </ActionButton>
        </View>
    );
  }

 buttonPressed() {
    this.props.navigation.navigate('NewCase', {});
  }

Answer 1

You don't execute buttonPressed at all. Fix it with ():

<ActionButton buttonColor="#1E73C1" onPress={() => this.buttonPressed()}>

Other way would be:

<ActionButton buttonColor="#1E73C1" onPress={this.buttonPressed.bind(this)}>

And like said in the comments, you should ensure that navigation actually exists in the props.