Question

I’m trying to create a simple username/password login screen. I have the layout done, and right now, I’m trying to set it so when the username (EditText) == "crete", then it should do something. Here is my code…:

public class Login extends Activity {
   public static EditText username, password;
   public Button loginbutton;
   boolean accessgranted;
   public String dbu, dbp, user1;

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

       username = (EditText) this.findViewById(R.id.username);
       password = (EditText) this.findViewById(R.id.password);        
       loginbutton = (Button) this.findViewById(R.id.loginbutton);

       user1 = "crete";

       loginbutton.setOnClickListener(new View.OnClickListener() {
         public void onClick(View v) {
            try{
               dbu = (username.getText()).toString();
            }
            finally{
               if (dbu == user1){
                   username.setText("SUCCESS");
               }
            }
         }
       });
    }
}   

this, sadly, doesn’t work. It correctly converts it to a string (i think) because when I tested this code out :

 loginbutton.setOnClickListener(new View.OnClickListener() {
     public void onClick(View v) {
        try{
              dbu = (username.getText()).toString();
        }
        finally{
                username.setText("done" + dbu);
            }
        }

     }
});

It correctly enters what you entered into the EditText, plus the word “done”.

There seems to be a problem with creating if-then statements??

Answer 1

You test for String equality with the method .equals("String").

With == you are testing if the references to the objects are equal.