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java.lang.ClassCastException: java.util.HashMap cannot be cast to java.lang.String

Hey can anyone help me out clearing this error :

java.lang.ClassCastException: java.util.HashMap cannot be cast to java.lang.String at in.xyz.firebase.MainActivity$1.onDataChange

MainActivity.java

import java.lang.String;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle; 
import android.view.Menu;
import android.view.MenuItem;
import android.widget.Button;
import android.widget.TextView;

import com.firebase.client.DataSnapshot;
import com.firebase.client.Firebase;
import com.firebase.client.FirebaseError;
import com.firebase.client.ValueEventListener;

public class MainActivity extends AppCompatActivity {

private Firebase mRef;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    Firebase.setAndroidContext(this);
}

@Override
protected void onStart() {
    super.onStart();
    Button b1 = (Button) findViewById(R.id.buttonSunny);
    Button b2 = (Button) findViewById(R.id.buttonFoggy);
   final  TextView t1 = (TextView) findViewById(R.id.textView);

    mRef = new Firebase("https://docs-    examples.firebaseio.com/web/saving-data/fireblog/posts");

    mRef.addValueEventListener(new ValueEventListener() {
        @Override
        public void onDataChange(DataSnapshot dataSnapshot) {
            String value = (String) dataSnapshot.getValue();
            t1.setText(value);
        }

        @Override
        public void onCancelled(FirebaseError firebaseError) {

        }
    });
}
}

Does anyone have any idea how to resolve this?

like image 379
Saikiran Sondarkar Avatar asked Jan 25 '26 10:01

Saikiran Sondarkar


1 Answers

Always use a POJO rather than a HashMap :)

Great answer by @Rahul Chaurasia! But I just wanted to expand upon how you return another class.

Let's say you have a class of Todo:

public class Todo {
  private String author;
  private String message;

  public Todo() {
    // empty default constructor, necessary for Firebase to be able to deserialize blog posts
  }

  public String getAuthor() {
    return author;
  }

  public String getMessage() {
    return message;
  }

}

Since this class has an empty constructor it's eligible for deserialization with Firebase.

  // Text view
  final TextView t1 = (TextView) findViewById(R.id.textView);

  // Get a reference to our posts
  Firebase ref = new Firebase("<my-firebase-db>/todos/1");

  // Get the todos
  ref.addValueEventListener(new ValueEventListener() {
      @Override
      public void onDataChange(DataSnapshot snapshot) {
         Todo todo = snapshot.getValue(Todo.class);
         t1.setText(todo.getMessage());
      }

      @Override
      public void onCancelled(FirebaseError firebaseError) {
          System.out.println("The read failed: " + firebaseError.getMessage());
      }
  });

Using a class rather than a HashMap provides much better type-safety. And, it takes advantage of something that's just built-in to the Firebase SDK.

like image 78
David East Avatar answered Jan 28 '26 00:01

David East



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