Android 搜索栏不工作
我在实时数据库的教师集合中存储了一些值。 然而,当我尝试使用我的搜索栏时,我的应用程序崩溃了 SETU方法Android 搜索栏不工作,android,firebase,firebase-realtime-database,Android,Firebase,Firebase Realtime Database,我在实时数据库的教师集合中存储了一些值。 然而,当我尝试使用我的搜索栏时,我的应用程序崩溃了 SETU方法 private void setAdapter(final String searchedstring) { databaseReference.child("teacher").addListenerForSingleValueEvent(new ValueEventListener() { @Override public void onDa
private void setAdapter(final String searchedstring) {
databaseReference.child("teacher").addListenerForSingleValueEvent(new ValueEventListener() {
@Override
public void onDataChange(@NonNull DataSnapshot dataSnapshot) {
fullnamelist.clear();
recyclerView.removeAllViews();
int counter=0;
for (DataSnapshot snapshot:dataSnapshot.getChildren()){
String uid =snapshot.getKey();
String full_name=snapshot.child("name").getValue(String.class);
if(full_name.toLowerCase().contains(searchedstring)){
fullnamelist.add(full_name);
counter++;
}
if(counter==10)
break;
}
searchAdapter = new searchadapter(getActivity(),fullnamelist);
recyclerView.setAdapter(searchAdapter);
}
searchadapter.java
public class searchadapter extends RecyclerView.Adapter<searchadapter.SearchViewHolder> {
Context context;
ArrayList<String> fullnamelist;
class SearchViewHolder extends RecyclerView.ViewHolder{
TextView full_name;
public SearchViewHolder(@NonNull View itemView) {
super(itemView);
full_name=(TextView)itemView.findViewById(R.id.full_name);
}
}
public searchadapter(Context context, ArrayList<String> fullnamelist) {
this.context = context;
this.fullnamelist = fullnamelist;
}
@NonNull
@Override
public searchadapter.SearchViewHolder onCreateViewHolder(@NonNull ViewGroup parent, int viewType) {
View view = LayoutInflater.from(context).inflate(R.layout.search_list, parent ,false);
return new searchadapter.SearchViewHolder(view);
}
@Override
public void onBindViewHolder(@NonNull SearchViewHolder holder, int position) {
holder.full_name.setText(fullnamelist.get(position));
}
@Override
public int getItemCount() {
return fullnamelist.size();
}
更改此选项:
String full_name = snapshot.child("name").getValue(String.class);
为此:
String full_name = snapshot.child("full_name").getValue(String.class);
根据您的数据库,您有一个名为
full\u name
的子属性,因此您需要将数据库中的确切名称作为方法child()
中的参数。您使用的键(“name”)无效,这会导致full\u name
变为null,因此出现null指针异常
试试这个
String full_name=snapshot.child("full_name").getValue(String.class);
full_name
为空,我已向其添加了值,请查看使用无效密钥的图像的链接。请查看我的答案。是的,现在它不会崩溃,但搜索栏不工作。您应该为此问题发布一个新问题,并附上相关代码,如方法代码setAdapter
String full_name = snapshot.child("full_name").getValue(String.class);
String full_name=snapshot.child("full_name").getValue(String.class);