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，我正在尝试插入到sqlite android，如下所示，它的工作原理。唯一的问题是我似乎失去了每个问题与所有答案之间的关系甚至widget_id。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，如代码所示：
question_id，question_title，answer_id，answer_value和widget_id 。
如何在sqlite android中 INSERTING 和 RETRIEVING 时，在每个问题，所有答案和小部件ID之间维护json对象中存在的关系。
修改
所以这是我得到的例外：
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)
和另一个
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)
修改 * 从mysql服务器下载的FragmentStatePagerActivity，将值传递给PagerAdapter然后加载片段 *
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;
}
}
正如您所看到的，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，你刚刚在第二个答案中向我展示。最后我在TableAnswers类和TableQuestions类中打开它因为我们正在将数据插入数据库。
答案 0 :(得分：1)
你必须创建两个表 1）质疑父母 2）应答
1）问题表字段：
auto_Id (primary Key) auto increment
question_id
description
widget_id
2）回答表字段：
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，此代码可用于将数据插入表...成功..
答案 1 :(得分：0)
您可以根据QuestionId制作模型类的对象。你的模型类将包含一个Answers的ArrayList，它将是另一个模型类。根据问题ID，您有一个对象可以获得该问题的所有答案。
可以通过a为每个循环确定。
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a - ＆gt;对于每个对象＆amp; QuestionModel中的答案列表 - ＆gt;与问题ID匹配并相应插入。
在检索时，您已将列表整理出来
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直接在您的类中映射响应，而不是像代码中的json解析，例如：
Gson gson = new Gson();
QuestionModel questionModel= new QuestionModel();
questionModel= gson.fromJson(responseContent,QuestionModel.class);
//where responseContent is your jsonString
然后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等注释。 （所以不需要使用Serializable）
答案 2 :(得分：0)
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;
}