Question

我正在尝试将数据写入firebase云。它似乎不起作用。你有什么建议？

$.when

Answer 1

import android.content.Intent;
import android.os.Bundle;
import android.support.v7.app.AppCompatActivity;
import android.widget.EditText;

import com.firebase.client.Firebase;

public class MainActivity extends AppCompatActivity {

    String muruga2,key="U10";
    Firebase ref;

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

    }


    public void additem() {

        EditText muruga = (EditText) findViewById(R.id.muruga);
        muruga2 = muruga.getText().toString();
        ref = new Firebase("https://project-7493703210.firebaseio.com/");
        postvalue(key,muruga2 )// key must be unique for every entry
        Intent inti = new Intent(MainActivity.this,NextActivity.class);
        startActivity(inti);
    }
}

private void postvalue(String key,  String value) {
        Post post = new Post(key,value);
        final Map<String, Object> postValues = post.toMapBodyStyle();
        ref.child("Name").setValue(postValues).addOnSuccessListener(new OnSuccessListener<Void>() {
                        @Override
                        public void onSuccess(Void aVoid) {
                            Toast.makeText(RegistrationActivity.this, "Data added successfully", Toast.LENGTH_SHORT).show();

                        }
                    }).addOnFailureListener(new OnFailureListener() {
                        @Override
                        public void onFailure(@NonNull Exception e) {
                            Toast.makeText(RegistrationActivity.this, "Error in adding user", Toast.LENGTH_SHORT).show();

                        }
                    });
    }

发布课程

 @IgnoreExtraProperties
public class Post {
    public String value;
    public String key;

    public Post() {
        // Default constructor required for calls to DataSnapshot.getValue(Post.class)
    }



    public Post(String key,String val) {
        this.value = val;
        this.key=key;
    }

    @Exclude
    public Map<String, Object> toMap() {
        HashMap<String, Object> result = new HashMap<>();
        result.put(key, val);
    }

    @Exclude
    public Map<String, Object> toMapBodyStyle() {
        HashMap<String, Object> result = new HashMap<>();
        result.put(key, value);
        return result;
    }

}

Firebase Android - 应用程序关闭，没有构建错误

1 个答案: