Firebase将数据保存为List，有时保存为Map对象

Question

我使用键将数据保存为数值，其中键是用户尝试的问题。如果用户尝试了所有问题，则可能是连续的。键可能是0,1,2,3 ......这是由Firebase保存为数组（如图像中所示）

JSON是

"parthgupta48@gmail,com": {
  "attemptedQuestions": [
    {
      "answer": "533",
      "status": "wrong",
      "submissionTime": 1487796702453
    },
    {
      "answer": "9",
      "score": 10,
      "status": "correct",
      "submissionTime": 1487878744867
    },
    {
      "answer": "4",
      "status": "wrong",
      "submissionTime": 1487956858227
    },
    {
      "answer": "3",
      "status": "wrong",
      "submissionTime": 1488056247303
    },
    null,
    {
      "answer": "25",
      "score": 16,
      "status": "correct",
      "submissionTime": 1488212417465
    },
    {
      "answer": "3",
      "status": "wrong",
      "submissionTime": 1488380348815
    },
    null,
    null,
    null,
    {
      "answer": "50",
      "score": 11,
      "status": "correct",
      "submissionTime": 1488648738615
    }
  ],

当键不连续时，例如0,9,15 ......当用户只尝试某些问题时会发生这种情况，然后Firebase将其保存为地图（如图像中所示）

JSON是

"priyanshu96,goyal@gmail,com" : {
      "attemptedQuestions" : {
        "0" : {
          "answer" : "667",
          "status" : "wrong",
          "submissionTime" : 1487773682189
        },
        "10" : {
          "answer" : "50",
          "submissionTime" : 1488646361162
        }
      },

这是我用来在Android中检索数据的代码。

try {
       mCurrentUser = child.getValue(UserObject.class);
    } catch (Exception e) {

       Utils.makeToast("Some Error Occurred ",LoginActivity.this);

   }

userObject类

public class UserObject implements Serializable {

private String name;
private String uid;
private String email;
private String photoUrl;
private String contact;//bekaar
private long credits;
public Map<String,AttemptedByUserObject> attemptedQuestions;

public void setAttemptedQuestions(Map<String, AttemptedByUserObject> attemptedQuestions) {
    this.attemptedQuestions = attemptedQuestions;
}

public Map<String, AttemptedByUserObject> getAttemptedQuestions() {
    return attemptedQuestions;
}

public long getCredits() {
    return credits;
}

public void setCredits(long credits) {
    this.credits = credits;
}

public String getName() {
    return name;
}

public void setName(String name) {
    this.name = name;
}

public String getUid() {
    return uid;
}

public void setUid(String uid) {
    this.uid = uid;
}

public String getEmail() {
    return email;
}

public void setEmail(String email) {
    this.email = email;
}

public String getPhotoUrl() {
    return photoUrl;
}

public void setPhotoUrl(String photoUrl) {
    this.photoUrl = photoUrl;
}

public String getContact() {
    return contact;
}

public void setContact(String contact) {
    this.contact = contact;
}}

但是，当数据保存为列表时，问题attemptedQuestions需要列出，当数据保存为Map时，问题需要Map。

我该如何解决这个问题？

Answer 1

当使用从零（即0、1、2等）开始的连续键保存信息时，您将返回一个列表，在我的情况下，我是从一个开始的，因此由于某种原因它使我回到了列表以null元素开头。当键不连续时，将返回一个Map。了解了这一点之后，我最终通过类型检查对其进行了修补，由于列表中的空值，我删除了第一个元素。

when (it) {
    is Map<*, *> -> doSomething(it.values.toList() as List<HashMap<String, Any>>)
    is List<*> -> doSomething((it.subList(1,it.size)) as List<HashMap<String, Any>>)
    is Any -> return
}

在示例中，doSomething仅接受List<Hashmap<String, Any>>个参数，但变量it可以是Map或List。