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Java Firebase查询未返回其匹配的值_Java_Android_Firebase_Firebase Realtime Database - Fatal编程技术网

Java Firebase查询未返回其匹配的值

Java Firebase查询未返回其匹配的值,java,android,firebase,firebase-realtime-database,Java,Android,Firebase,Firebase Realtime Database,我的密码是 ValueEventListener valueEventListener = new ValueEventListener() { @Override public void onDataChange(@NonNull DataSnapshot dataSnapshot) { for(DataSnapshot single : dataSnapshot.getChildren()){ quick

我的密码是

ValueEventListener valueEventListener = new ValueEventListener() {
        @Override
        public void onDataChange(@NonNull DataSnapshot dataSnapshot) {
            for(DataSnapshot single : dataSnapshot.getChildren()){
                quickResultsAdapter.add(single.child(getString(R.string.db_displayName)).getValue().toString());
            }
        }

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

        }
    };
查询是这样的

query =  application.getMRef().child(getString(R.string.db_user_details)).orderByChild(getString(R.string.db_displayName))
                    .startAt("vikass").endAt("vikass"+"\uf8ff");
                query.addValueEventListener(valueEventListener);

当我查询vikass时,如果我的数据库值是Vikassarma,它将不返回任何内容

rootNode
  user_details
       displayName: vikassharma
然后,您不需要添加childgetStringR.string.db\u用户\u详细信息,只需执行以下操作:

query =  application.getMRef().orderByChild(getString(R.string.db_displayName))
                .startAt("vikass").endAt("vikass"+"\uf8ff");
            query.addValueEventListener(valueEventListener);

你能给我们看一下你的数据库结构吗?user_details{displayName:Vikassarma}是string.db_user_details与你数据库中孩子的名字匹配吗?是的,我有200%的把握