Firebase 实时数据库 - 检索数据 Android Studio

Question

Getting error: com.google.firebase.database.DatabaseException: Can't convert object of type java.lang.String to type com.example.ken.careerapp.Models.Jobs

Using a fragment below:

public View onCreateView(@NonNull LayoutInflater inflater, @Nullable ViewGroup container, @Nullable Bundle savedInstanceState) {
        View view = inflater.inflate(R.layout.job_listings_fragment, container, false);
        recyclerView = view.findViewById(R.id.recyclerviewjob);
        recyclerView.setHasFixedSize(true);
        recyclerView.setLayoutManager(new LinearLayoutManager(getContext()));
        recyclerAdapterJob = new RecyclerAdapterJob(jobs,JobFragment.this::onJobClick);
        recyclerView.setAdapter(recyclerAdapterJob);
        jobs = new ArrayList<Jobs>();
        //initData();
        databaseReference = FirebaseDatabase.getInstance().getReference().child("Jobs");

        databaseReference.addListenerForSingleValueEvent(new ValueEventListener() {
            @Override
            public void onDataChange(@NonNull DataSnapshot dataSnapshot) {
                for(DataSnapshot ds: dataSnapshot.getChildren()){
                    Jobs data = ds.getValue(Jobs.class);
                    jobs.add(data);
                }
                recyclerAdapterJob.notifyDataSetChanged();
            }

            @Override
            public void onCancelled(@NonNull DatabaseError databaseError) {

            }
        });
        //recyclerView.setAdapter(new RecyclerViewAdapter2(initData()));

        return view;
    }

The job model class which is passed to getValue in datasnapshot above:

public class Jobs {
private String description;
private String location;
private String deadline;

public Jobs() {
}

public Jobs(String description, String location, String deadline) {
    this.description = description;
    this.location = location;
    this.deadline = deadline;
}

public String getDescription() {
    return description;
}

public void setDescription(String description) {
    this.description = description;
}

public String getLocation() {
    return location;
}

public void setLocation(String location) {
    this.location = location;
}

public String getDeadline() {
    return deadline;
}

public void setDeadline(String deadline) {
    this.deadline = deadline;
}

}

FirebaseRealtime database

Answer 1

Give a photo or your realtime database architecture. that will help much to ans this issue.

Answer 2

You are getting the following error:

Getting error: com.google.firebase.database.DatabaseException: Can't convert object of type java.lang.String to type com.example.ken.careerapp.Models.Jobs

Because you are trying to convert a String, into an object of type Jobs, and this actually not possible in Java. As I see in your screenshot, under the Jobs node, there is only a single Jobs object that contains exactly three String properties. So when you iterate that node, you get String objects back and not Jobs objects, hence that error.

To solve this, you should create multiple objects under that node, as I understand, you want to display them in a RecyclerView. To solve this, you should use the push() method, when you add those jobs to the database. You didn't share that part of code, but I imagine that looks similar to this:

databaseReference = FirebaseDatabase.getInstance().getReference().child("Jobs");
databaseReference.setValue(jobs);

In which jobs is an object of type Jobs. This is not correct, as it produces that error when reading data. To solve this, please change the above line of code to:

databaseReference.push().setValue(jobs);

In this case, your new structure will look like this:

Firebase-root
  |
  --- Jobs
       |
       --- pushedId
             |
             ---deadline: "September"
             |
             ---description: "Intern Software Developer"
             |
             ---location: "Moi Avenue Street"

And the rest of the code should work.

PS

Don't ignore potential errors!

@Override
public void onCancelled(@NonNull DatabaseError databaseError) {
    Log.d("TAG", databaseError.getMessage());
}

Always log the error message.