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Android Build an Interactive Story App (Retired) Finishing the User Interface Using a Model in the Controller

Vidit Shah
Vidit Shah
6,037 Points

New space Ship task

Compile error

LandingActivity.java
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;

public class LandingActivity extends Activity {

    public Button mThrustButton;
    public TextView mTypeLabel;
    public EditText mPassengersField;

    public Spaceship mSpaceship;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_landing);

        mThrustButton = (Button)findViewById(R.id.thrustButton);
        mTypeLabel = (TextView)findViewById(R.id.typeTextView);
        mPassengersField = (EditText)findViewById(R.id.passengersEditText);

        // Add your code here!
      mSpaceship = new Spaceship(String s1)
      {
        s1="FIRELY";
      }
    }
}
Spaceship.java
public class Spaceship {
    private String mType;
    private int mNumPassengers = 0;

    public String getType() {
      return mType;
    }

    public void setType(String type) {
      mType = type;
    }

    public int getNumPassengers() {
      return mNumPassengers;
    }

    public void setNumPassengers(int numPassengers) {
      mNumPassengers = numPassengers;
    }

    public Spaceship() {
      mType = "SHUTTLE";
    }

    public Spaceship(String type) {
      mType = type;
    }
}

1 Answer

Luis Cole
Luis Cole
17,228 Points

Your error is right here:

mSpaceship = new Spaceship(String s1)
      {
        s1="FIRELY";
      }

You only need to type "FIRELY" as the argument between the parenthesis.

mSpaceship = new Spaceship("FIREFLY");

You can also declare a String like you did an set equal to "FIRELY" and then put that String as the argument

String firefly = "FIREFLY";
mSpaceship = new Spaceship(firefly);