如何从Android中保存Firebase数据库中的数据数组？

Question

我是Firebase数据库的新手,我只想保存Firebase数据库中的数据数组.当我试图保存它显示错误.下面我在保存数据时提到了POJO类和Logcat错误.

POJO类编码

@IgnoreExtraProperties
public class StudentReviews {

public String student_name;
public String student_class;
public String student_image;
public String student_section;

public ArrayList<Reviews> reviewsList = new ArrayList<>();

public void StudentReviews() {}

public class Reviews {
        public String subject;
        public String marks_taken;
        public String total_marks;
        public String teacher;
    }
}

保存数据

StudentReviews studentReviews = new StudentReviews();
studentReviews.student_name = et_studentName.getText().toString();
studentReviews.student_class = et_student_class.getText().toString();
studentReviews.student_image = "";
studentReviews.student_section = sp_section.getSelectedItem().toString();
studentReviews.reviewsList = reviewsArray;

dbUploadReview.push().setValue(studentReviews); //165th Line in Logcat Error

Logcat错误

com.google.firebase.database.DatabaseException: Invalid key: this$0. Keys must not contain '/', '.', '#', '$', '[', or ']'
at com.google.android.gms.internal.zzbtf.zzjq(Unknown Source)   
at com.google.android.gms.internal.zzbtf.zzay(Unknown Source)
at com.google.android.gms.internal.zzbtf.zzay(Unknown Source)
at com.google.android.gms.internal.zzbtf.zzay(Unknown Source)  
at com.google.firebase.database.DatabaseReference.zza(Unknown Source)
at com.google.firebase.database.DatabaseReference.setValue(Unknown Source)
at com.trending.trendingadmin.fragment.WriteReview.uploadReviews(WriteReview.java:165)

当我 在Firebsae数据库中保存的POJO类数据中完全删除了这个public ArrayList<Reviews> reviewsList = new ArrayList<>();reviewsList .

任何人都知道帮我解决这个问题.

更新

我通过两种方式解决了这个问题.OlegOsipenko给出的建议

1)使内部类成为静态

2)将内部类移动到另一个文件.

现在它的工作和我预期的一样好.

Answer 1

.setValue() 方法需要一个列表而不是数组。您可以使用对象列表。

Firebase ref = new Firebase("<my-firebase-app>/names"):
String[] names = {"John","Tim","Sam","Ben"};
List nameList = new ArrayList<String>(Arrays.asList(names));
// Now set value with new nameList
ref.setValue(nameList);

阅读 Firebase 文档以获取更多信息。