遍历嵌套的json对象并将数据保存在android sqlite

Question

我有一个嵌套的 JSON 对象，如下所示：

[
   {
  "question_id":"1",
  "description":"What is your gender ?",
  "widget_id":"1",
  "answers":[
     {
        "answer_text":"Male",
        "answer_id":"1"
     },
     {
        "answer_text":"Female",
        "answer_id":"2"
     }
  ]
   },
   {
  "question_id":"2",
  "description":"Which animal best describes your personality ?",
  "widget_id":"2",
  "answers":[
     {
        "answer_text":"Cat",
        "answer_id":"3"
     },
     {
        "answer_text":"Horse",
        "answer_id":"4"
     },
     {
        "answer_text":"Dove",
        "answer_id":"5"
     },
     {
        "answer_text":"Lion",
    "answer_id":"6"
    },
   {
            "answer_text":"Chameleon",
            "answer_id":"7"
     }
  ]
   },
   {
      "question_id":"3",
  "description":"Do you like meeting other people ?",
  "widget_id":"3",
  "answers":[

  ]
   },
   {
  "question_id":"4",
  "description":"On a scale of 1-10, how would you rate your sense of humour ?",
  "widget_id":"4",
  "answers":[

  ]
   },
   {
  "question_id":"5",
  "description":"Are you afraid of the dark ?",
  "widget_id":"1",
  "answers":[
     {
        "answer_text":"No",
        "answer_id":"8"
     },
     {
        "answer_text":"Yes",
        "answer_id":"9"
     }
  ]
   },
   {
  "question_id":"6",
  "description":"Is it true that cannibals do not eat clowns because they taste kind of funny ?",
  "widget_id":"3",
  "answers":[

  ]
   },
   {
  "question_id":"7",
  "description":"What is your email address ? (Optional)",
  "widget_id":"3",
  "answers":[

  ]
   }
]

从mysql服务器检索后，我试图插入sqlite android，如下所示，它可以工作。唯一的问题是我似乎失去了每个问题与其所有答案甚至widget_id之间的关系。因为一些问题有多个答案选项。

JSONArray aJson = new JSONArray(sJson);
ArrayList<Question> Question_Id_array = new ArrayList<Question>();

            for (int i = 0; i < aJson.length(); i++) {
                JSONObject json = aJson.getJSONObject(i);

                Question que = new Question();

                Question id = new Question();

                que.setDescription(json.getString("description"));

                id.setQuestionId(Integer.parseInt(json
                        .getString("question_id")));
                que.setWidgetId((Integer.parseInt(json
                        .getString("widget_id"))));
JSONArray cJson = json.getJSONArray("answers");
                ArrayList<Answer> ans = que.getAnswers();

                for (int k = 0; k < cJson.length(); k++) {
                    JSONObject Objectjson = cJson.getJSONObject(k);
                    Answer answer = new Answer();

                    answer.setAnswer_Text(Objectjson
                            .getString("answer_text"));
                    answer.setAnswer_Id(Integer.parseInt(Objectjson
                            .getString("answer_id")));
ans.add(answer);

String answer_value = answer.getAnswer_Text()
                            .toString();

                    int answer_id = answer.getAnswer_Id();

                    String question_title = que.getDescription().toString();

                    int question_id = que.getQuestionId();

                    int widget_id = que.getWidgetId();

                    ContentValues cv = new ContentValues();
                    cv.put(ResponseDetails.KEY_QUESTION_ID,question_id);
                    cv.put(ResponseDetails.KEY_QUESTION_DESCRIPTION,question_title);
                    cv.put(ResponseDetails.ANSWER_ID, answer_id);
                    cv.put(ResponseDetails.KEY_ANSWER_VALUE,answer_value);
                    cv.put(ResponseDetails.WIDGET_ID, widget_id);


                    getApplicationContext().getContentResolver()

                    .insert(ResponseContentProvider.CONTENT_URI2, cv);

                }

我目前有一个包含所有列的表，如代码中所示：

question_id、question_title、answer_id、answer_value 和 widget_id。

如何维护每个问题之间的 json 对象中存在的关系，所有问题的答案和小部件 ID 都同时 INSERTING 和 RETRIEVING 从 sqlite android。 p>

编辑

所以这是我现在得到的例外：

02-11 15:44:33.487: E/AndroidRuntime(1336): FATAL EXCEPTION: main
02-11 15:44:33.487: E/AndroidRuntime(1336): java.lang.NullPointerException
02-11 15:44:33.487: E/AndroidRuntime(1336):     at         com.mabongar.survey.TableAnswers.insert(TableAnswers.java:53)
02-11 15:44:33.487: E/AndroidRuntime(1336):     at     com.mabongar.survey.FragmentStatePagerActivity$FetchQuestions.onPostExecute(FragmentStatePagerActivity.java:    177)
02-11 15:44:33.487: E/AndroidRuntime(1336):     at    com.mabongar.survey.FragmentStatePagerActivity$FetchQuestions.onPostExecute(FragmentStatePagerActivity.java:    1)
02-11 15:44:33.487: E/AndroidRuntime(1336):     at android.os.AsyncTask.finish(AsyncTask.java:631)
02-11 15:44:33.487: E/AndroidRuntime(1336):     at     android.os.AsyncTask.access$600(AsyncTask.java:177)
02-11 15:44:33.487: E/AndroidRuntime(1336):     at   android.os.AsyncTask$InternalHandler.handleMessage(AsyncTask.java:644)
02-11 15:44:33.487: E/AndroidRuntime(1336):     at   android.os.Handler.dispatchMessage(Handler.java:99)

还有一个

02-11 15:44:39.867: E/SQLiteLog(1357): (14) cannot open file at line 30191 of [00bb9c9ce4]
02-11 15:44:39.867: E/SQLiteLog(1357): (14) os_unix.c:30191: (2)   open(/data/data/com.mabongar.survey/databases/responsetable.db) - 
02-11 15:44:40.017: E/SQLiteDatabase(1357): Failed to open database    '/data/data/com.mabongar.survey/databases/responsetable.db'.
02-11 15:44:40.017: E/SQLiteDatabase(1357): android.database.sqlite.SQLiteCantOpenDatabaseException:   unknown error (code 14): Could not open database
02-11 15:44:40.017: E/SQLiteDatabase(1357):     at   android.database.sqlite.SQLiteConnection.nativeOpen(Native Method)

编辑 *从 mysql 服务器下载的 FragmentStatePagerActivity，将值传递给 PagerAdapter，然后加载 Fragments*

public class FragmentStatePagerActivity extends ActionBarActivity {

public SQLiteDatabase db;
private final String DB_PATH = "/data/data/com.mabongar.survey/databases/";

private static final String DATABASE_NAME = "responsetable.db";
// AsyncTask Class

private class FetchQuestions extends AsyncTask<String, Void, String> {

    @SuppressWarnings("static-access")
    @Override
    protected String doInBackground(String... params) {

        if (params == null)

            return null;

        try{

            String mPath = DB_PATH + DATABASE_NAME;

            db = SQLiteDatabase.openDatabase(mPath, null, SQLiteDatabase.CONFLICT_NONE);

        }catch(SQLException e){

            Log.e("Error ", "while opening database");
            e.printStackTrace();
        }

//          // get url from params

        String url = params[0];

        try {
            // create http connection
            HttpClient client = new DefaultHttpClient();
            HttpGet httpget = new HttpGet(url);

            // connect
            HttpResponse response = client.execute(httpget);

            // get response
            HttpEntity entity = response.getEntity();

            if (entity == null) {
                return null;
            }

            // we get response content and convert it to json string
            InputStream is = entity.getContent();
            return streamToString(is);
        } catch (IOException e) {
            Log.e("Log message", "No network connection");
        }

        return null;
    }

}

如您所见，这就是我在 doInBackground() 方法中打开它的方式 然后我还在 pagerAdapter 类中打开它，因为它具有您刚刚在第二个答案中向我展示的 public ArrayList SelectAll() 方法。最后我在 TableAnswers 类和 TableQuestions 类中将其打开为好吧，因为我们正在向数据库中插入数据。

Answer 1

你必须创建两个表 1）问题父母 2）回答

1)问题表字段：

    auto_Id (primary Key) auto increment
    question_id
    description
    widget_id

2)答案表字段：

   auto_Id (primary Key) auto increment
   answer_text
   answer_id
   question_id

public void table_question{
//this functin is used for insert data .......when pass data from json
public void insert(Arraylist<Model_question> modelArrlist){

     for (Model_question model : modelArrlist) {
    ContentValues values = new ContentValues();
        values.put(auto_Id, model.auto_Id);
        values.put(question_id, model.question_id);
        values.put(description, model.description);
        values.put(widget_id, model.widget_id);
        sqldb.insert(TableName, null, values);

        for(Model_answer model_answer :model.arrAnswerList)
        {
           model_answer.question_id=model.question_id
          Tbl_answer.insert(model_master);
        }
     }

}
}

//这是tbl_answer插入方法

public class tbl_answer{

public void insert(Model_answer model_answer){

    ContentValues values = new ContentValues();
        values.put(auto_Id, model.auto_Id);
        values.put(question_id, model.question_id);
        values.put(answer_text, model.answer_text);
      values.put(answer_id, model.answer_id);

      }
}


public void Model_question {

    public String question_id,
        description,
        widget_id;
   public List<Model_answer> arrAnswerList=new ArrayList<Model_answer>;
}

public void Model_answer{

    public String answer_text,
       answer_id,
       question_id;

 }

请检查此代码，此代码将有助于将数据插入到两个表中..成功..

Answer 2

您可以根据 QuestionId 创建模型类的对象。您的模型类将包含一个 ArrayList of Answers，这将是另一个模型类。根据问题 ID，您有一个对象可以包含该问题的所有答案。

这可以通过 for each 循环来确定。

class QuestionModel {
  String questionId;
  String description;
  String widgetId;
  ArrayList<Answers> answers;
  //getter setters here
}  

class AnswersModel{
 String answerText;
 String answerId;
//getter setter
}  

插入时，为每个循环使用一个 -> 为 QuestionModel 中的每个对象和答案列表 -> 匹配问题 ID 并相应地插入。
检索时，您已整理好列表。
http://docs.oracle.com/javase/1.5.0/docs/guide/language/foreach.html
How does the Java 'for each' loop work?

此外，您可以使用 Gson 并将响应直接映射到您的类，而不是像在您的代码中那样解析 json，例如：

Gson gson = new Gson();
QuestionModel questionModel= new QuestionModel();
questionModel= gson.fromJson(responseContent,QuestionModel.class); 
//where responseContent is your jsonString  

那么您不必单独处理解析或检查 Answer 数组。
检查：https://code.google.com/p/google-gson/

对于命名差异（根据webservice中的变量），可以使用@SerializedName等注解。 （所以不需要使用Serializable）

Answer 3

public static ArrayList<Model_question> SelectAll() 
    ArrayList<Model_question> arrModelList = null;
        Cursor cursor = null;
        String Query = "Select * from " + TableName;
        cursor = sqldb.rawQuery(Query, null);
        if (cursor != null && cursor.moveToFirst()) {
            arrModelList = new ArrayList<Model_question>();
            do {
                Model_question model = new Model_question();
                        model.auto_Id= (cursor.getString(cursor.getColumnIndex("auto_Id")));

                model.question_id= (cursor.getString(cursor.getColumnIndex("question_id")));
                model.description= (cursor.getString(cursor
                        .getColumnIndex("description")));
                model.widget_id= (cursor.getString(cursor.getColumnIndex("widget_id")));
                        model.arrAnswerList= Tbl_answer
                        .selectIdWiseData(model.question_id);
                arrModelList.add(model);
            } while (cursor.moveToNext());
            cursor.close();
        }// end if(cursor!=null)
        return arrModelList;
    }
}

public void selectIdWiseData(String question_id){

ArrayList<Model_answer> arrayList = null;
        Log.d("tag", "Model_answerId" + inId);
        String Query = "Select * from " + TableName
                + " where question_id='" + question_id+ "'";
        Log.d("tag", "Model_answer Query" + Query);
        Cursor cursor = sqldb.rawQuery(Query, null);
        if (cursor != null && cursor.moveToFirst()) {
            arrayList = new ArrayList<Model_answer>();
            do {
                Model_answer model = new Model_answer();
                model.autoId = (cursor.getString(cursor.getColumnIndex(AUOTID)));
                model.question_id= (cursor.getString(cursor
                        .getColumnIndex("question_id")));
                model.answer_text= (cursor.getString(cursor
                        .getColumnIndex("answer_text")));
model.answer_id= (cursor.getString(cursor
                        .getColumnIndex("answer_id")));


                arrayList.add(model);
            } while (cursor.moveToNext());
            cursor.close();
        }// end if(cursor!=null)
        return arrayList;
}