android - 遍历嵌套的 json 对象并将数据保存在 android sqlite 中

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，我尝试插入到 sqlite android 中，如下所示，它有效。唯一的问题是我似乎失去了每个问题与所有答案甚至 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);

                }

我目前有一个表，其中包含代码中所见的所有列:

question_id、question_title、answer_id、answer_value 和 widget_id。

我如何在来自 sqlite android 的 INSERTING 和 RETRIEVING 中保持每个问题之间存在于 json 对象中的关系，所有问题的答案和小部件 ID。

编辑

所以这是我现在得到的异常:

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 fragment *

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也在 pagerAdapter 类中打开它，因为它有 public ArrayList SelectAll() 方法，你刚刚在你的第二个答案中向我展示了这个方法。最后我在 TableAnswers 类和 TableQuestions 类中打开它作为好吧，因为我们要将数据插入数据库。

Answer 1

你必须创建两个表1)询问 parent 2)回答

1)问题表字段:

    auto_Id (primary Key) auto increment
    question_id
    description
    widget_id

2)answer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..