android - Firebase 将数据保存为列表，有时保存为 map 对象

Question

我将带有键的数据保存为数值，其中键是用户尝试的问题。如果用户尝试了所有问题，例如，它可能是连续的。键可能是 0,1,2,3... 这由 Firebase 保存为数组(如图所示)



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 ......当用户仅尝试某些问题时会发生这种情况，然后 Firebase 将其保存为 map (如图所示)



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);

   }

用户对象类

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;
}}

但问题是 attemptedQuestions数据保存为列表时需要列表，需要为 Map当数据保存为 Map .

我怎么解决这个问题？

Answer 1

当使用从零开始的连续键(即 0、1、2 等)保存信息时，您将返回一个列表，在我的情况下，我是从 One 开始的，所以出于某种原因，它让我返回了一个以 a 开头的列表空元素。当键不连续时，将返回一个 Map。在了解了这一点后，我最终通过类型检查对其进行了修补，并且由于列表案例中的 null ，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 只需要 List<Hashmap<String, Any>>参数，但变量 it可以是 map 或列表。