Question

    import com.google.firebase.database.DataSnapshot;
   import com.google.firebase.database.DatabaseError;
   import com.google.firebase.database.DatabaseReference;
   import com.google.firebase.database.FirebaseDatabase;
   import com.google.firebase.database.ValueEventListener;
   public class PlayId  {
private DatabaseReference mDatabase = FirebaseDatabase.getInstance().getReference();
public String id;
public String name;
public String getId() {
    return id;
}
public void setId(String id) {
    this.id = id;
}
public String getName() {
    return name;
}
public void setName(String name) {
    this.name = name;
}
public PlayId() {
}
public PlayId(String id, String name) {
    this.id = id;
    this.name = name;
}
public void writeNewPlayList(String userId,PlayId user) {        mDatabase.child("playlist").child("krikor").child(userId).setValue(user);
}
public void read()
{
    ValueEventListener postListener = new ValueEventListener() {
        @Override
        public void onDataChange(DataSnapshot dataSnapshot) {
            // Get Post object and use the values to update the UI
            PlayId post = dataSnapshot.getValue(PlayId.class);
            System.out.println(post.getId()+" tatatat "+post.name);
           }
            @Override
        public void onCancelled(DatabaseError databaseError) {
        }
    };
    mDatabase.addValueEventListener(postListener);
}}

简单地说，当我第一次这样做时。它首先工作，它存储播放列表 - ＆gt; Krikor-＆gt; playlistid-＆gt; playlistId / playlistName。然而，当我再次调用它时，它会更新数据。我只是希望它存储新数据播放列表 - ＆gt; Krikor-＆gt;新数据 - ＆gt;新数据/新数据。什么是解决方案？

        PlayId xs = new PlayId(""+playlistId,playlistName);
        xs.writeNewPlayList(xs.getId(),xs);

我有第二个问题，如何读回所有存储的值playlist-＆gt; krikor。因为我可能有播放列表 - ＆gt;詹姆斯。

Answer 1

使用此代码读取值

    DatabaseReference databaseReference1 = firebaseDatabase.getReference().child(KEY_METAINGREDIENT)
            .child(KEY_META_BURGER);
    final List<Ingredient> burger = new ArrayList<>();

    databaseReference1.addValueEventListener(new ValueEventListener() {
        @Override
        public void onDataChange(com.google.firebase.database.DataSnapshot dataSnapshot) {
            burger.clear();
            for (DataSnapshot dataSnapshot1 : dataSnapshot.getChildren()) {
                Ingredient ingredient = dataSnapshot1.getValue(Ingredient.class);
                burger.add(ingredient);
            }
        }

        @Override
        public void onCancelled(DatabaseError databaseError) {

        }
    });

Firebase无法正常工作

1 个答案: