java - Android/FireStore QuerySnapshot转换为CustomObject

我目前正在测试QuizApp中进行编程。游戏玩法非常简单，我只想要一个在线问题数据库，并且用户可以回答这些问题。
这是数据库的样子:


该集合问题包含一个唯一的ID和一个名为“content”的自定义对象(questionObject)。该数字只是我可以查询/搜索的简单内容。
这是我的questionAdder和查询UI。这只是一个小型测试应用程序。
公共(public)类questionAdder扩展了AppCompatActivity {

EditText pQuestion, pAnwerA, pAnswerB, pAnswerC, pAnswerD, number;
Button pAdd, query;
private DatabaseReference databaseReference;
private FirebaseFirestore firebaseFirestore;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.addquestion);

    firebaseFirestore = FirebaseFirestore.getInstance();

    pQuestion = (EditText) findViewById(R.id.question);
    pAnwerA = (EditText) findViewById(R.id.answerA);
    pAnswerB = (EditText) findViewById(R.id.answerB);
    pAnswerC = (EditText) findViewById(R.id.answerC);
    pAnswerD = (EditText) findViewById(R.id.answerD);
    number = (EditText) findViewById(R.id.number);

    pAdd = (Button) findViewById(R.id.addQuestion);
    pAdd.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            readQuestionStore();
        }
    });

    query = (Button) findViewById(R.id.query);
    query.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            CollectionReference questionRef = firebaseFirestore.collection("questions");
            questionRef.whereEqualTo("content.number", "20").get().addOnSuccessListener(new OnSuccessListener<QuerySnapshot>() {
                @Override
                public void onSuccess(QuerySnapshot queryDocumentSnapshots) {
                    questionObject pContent = queryDocumentSnapshots.toObjects(questionObject.class);
                }
            });
        }
    });

}

public void readQuestionStore(){

    Map<String, Object> pContent = new HashMap<>();
    pContent.put("question", pQuestion.getText().toString());
    pContent.put("Corr Answer", pAnwerA.getText().toString());
    pContent.put("AnswerB", pAnswerB.getText().toString());
    pContent.put("AnswerC", pAnswerC.getText().toString());
    pContent.put("AnswerD", pAnswerD.getText().toString());
    questionObject content = new questionObject(pContent, number.getText().toString()); //document("Essen").collection("Katalog")

    firebaseFirestore.collection("questions").add(content).addOnSuccessListener(new OnSuccessListener<DocumentReference>() {
        @Override
        public void onSuccess(DocumentReference documentReference) {
            Toast.makeText(questionAdder.this, "Klappt", Toast.LENGTH_LONG).show();
        }
    }).addOnFailureListener(new OnFailureListener() {
        @Override
        public void onFailure(@NonNull Exception e) {
            Toast.makeText(questionAdder.this, "Klappt nicht", Toast.LENGTH_LONG).show();
        }
    });
}

}

这就是我的questionObject的样子:

公共(public)类问题对象{

private Map<String, Object> content;
private String number;

public questionObject(){

}

public questionObject(Map<String, Object> pContent, String pNumber) {
    this.content = pContent;
    this.number = pNumber;
}

public Map<String, Object> getContent() {
    return content;
}

public void setContent(Map<String, Object> content) {
    this.content = content;
}

public String getNumber() {
    return number;
}

public void setNumber(String number) {
    this.number = number;
}

}

问题在onClickListener的那个questionAdder类中，我收到“不兼容的类型”错误(找到:java.utils.list必需:questionObject)。

query.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View view) {
                CollectionReference questionRef = firebaseFirestore.collection("questions");
                questionRef.whereEqualTo("content.number", "20").get().addOnSuccessListener(new OnSuccessListener<QuerySnapshot>() {
                    @Override
                    public void onSuccess(QuerySnapshot queryDocumentSnapshots) {
                        questionObject pContent = queryDocumentSnapshots.toObjects(questionObject.class);
                    }
                });
            }
        });

如果我将其更改为列表，则为空。因此，实际的问题是，如何使用数据库将CustomObject放入代码中。谢谢!

最佳答案

之所以会出现此错误，是因为QuerySnapshot是一种“包含”多个文档的类型。 Firestore不会为您决定是否要返回一堆对象或仅返回一个对象。
这就是为什么您可以采用两种不同的方法的原因:

将数据放入custom object的列表中:

List<questionObject> questionsList=new ArrayList<>();

if (!documentSnapshots.isEmpty()){
    for (DocumentSnapshot snapshot:queryDocumentSnapshots)
        questionsList.add(snapshot.toObject(questionObject.class));

}

如果您确定只会得到一个查询对象，则只需从返回的queryDocumentSnapshots中获取第一个对象:

questionObject object=queryDocumentSnapshots.getDocuments().get(0).toObject(questionObject.class);

您还应该注意以下几点:

为什么您要编写content.number而不是number？
似乎number是问题document中的分隔字段，因此您的代码应如下所示:

CollectionReference questionRef = firebaseFirestore.collection("questions");
questionRef.whereEqualTo("number", "20").get().addOnSuccessListener(new OnSuccessListener<QuerySnapshot>() {
    @Override
    public void onSuccess(QuerySnapshot queryDocumentSnapshots) {
        questionObject pContent = queryDocumentSnapshots.toObjects(questionObject.class);
    }
});

另外，请尝试将number字段更改为int，因为它不是String而是一个数字。

顺便说一句，在类的名称开头加一个大写字母比较容易接受，例如:QuestionObject question=new QuestionObject();