Question

我目前正在测试QuizApp。游戏玩法非常简单我只想要一个问题的在线数据库，用户可以回答它们。

这就是数据库的样子：

该集合问题包含唯一ID和名为＆＃39; content＆＃39;的自定义对象（questionObject）。这个数字只是我可以查询/搜索的简单内容。

这是我的问题添加和查询UI。它只是一个小测试应用程序。

public class 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：

public class questionObject {

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中的那个问题加法器类中，我收到了一个＆＃34;不兼容的类型＆＃34;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List，则为空。所以实际的问题是，如何使用数据库将CustomObject放入我的代码中。谢谢！

Answer 1

您收到此错误的原因是因为QuerySnapshot是一种＆＃34;包含＆＃34; 多个文件。 Firestore不会决定是否有一堆对象要返回，或者只返回一个。这就是为什么你可以采取两种不同的方法：

将数据放入custom object列表中：

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

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

}

如果您确定只能获得一个查询对象，则可以从返回的queryDocumentSnapshots获取第一个对象：
```
questionObject object=queryDocumentSnapshots.getDocuments().get(0).toObject(questionObject.class);
```

你应该注意的一些事情：

为什么要写content.number而不只是number？似乎number在您的问题document中是一个单独的字段，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);
    }
});

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

顺便说一下，写课程更容易接受。一开头有大写字母的名称，例如：QuestionObject question=new QuestionObject();

Android / FireStore QuerySnapshot转换为CustomObject

1 个答案: