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5,732 PointsLandingActivity interacts with the Spaceship model object we created. Start by setting spaceship to a new Spaceship
Couldnt solve this
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
public class LandingActivity extends AppCompatActivity {
public Button thrustButton;
public TextView typeLabel;
public EditText passengersField;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_landing);
thrustButton = (Button)findViewById(R.id.thrustButton);
typeLabel = (TextView)findViewById(R.id.typeTextView);
passengersField = (EditText)findViewById(R.id.passengersEditText);
// Add your code here!
Spaceship spaceship= new Spaceship("FIREFLY");
}
}
public class Spaceship {
private String shipType;
private int numPassengers = 0;
public String getShipType() {
return shipType;
}
public void setShipType(String shipType) {
this.shipType = shipType;
}
public int getNumPassengers() {
return numPassengers;
}
public void setNumPassengers(int numPassengers) {
this.numPassengers = numPassengers;
}
public Spaceship() {
shipType = "SHUTTLE";
}
public Spaceship(String shipType) {
this.shipType = shipType;
}
}
2 Answers
Steve Hunter
57,712 PointsHi there,
This is a simple fix. Because the variable called spaceship
has already been defined as being of type Spaceship
you don't need to take that step again. So, delete the first Spaceship
datatype.
Otherwise, you're declaring a new spaceship inside onCreate
. This will exist only in onCreate
so when the challenge tests the value of the member variable of the class, it'll not be set to anything.
I hope that makes sense!
Change your line of code to:
spaceship = new Spaceship("FIREFLY");
And that'll work fine.
Steve.
Sena Sari
5,732 PointsThanks for the answer it worked out well