迭代嵌套的json对象并在android sqlite中保存数据
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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sqlite android,如下所示,并且它的工作原理.唯一的问题是我似乎失去了每个问题和所有答案之间的关系甚至widget_id.因为有些问题有更多比一个答案选项.
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.
在从sqlite android进行INSERTING和RETRIEVING时,如何在每个问题,所有答案和小部件ID之间维护json对象中存在的关系.
编辑
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)
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;
}
}
如您所见,这就是我在doInBackground()方法中打开它的方法
然后我也在pagerAdapter类中打开它,因为它有公共的ArrayList SelectAll()方法,你刚才给我看了第二个答案.最后我在TableAnswers类和TableQuestions类中打开它,因为我们正在插入数据进入数据库.
解决方法:
你必须创建两个表
1)质疑父母
2)答案
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,此代码将有用于将数据插入两个表…成功..
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