遍历嵌套的json对象并将数据保存在android sqlite
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【中文标题】遍历嵌套的json对象并将数据保存在android sqlite【英文标题】：iterate through nested json object and save data in android sqlite 【发布时间】：2014-03-07 15:10:57 【问题描述】：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。
【问题讨论】：
【参考方案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..
【讨论】：
嗨@dipali 非常感谢您的解释。我正在立即对我的代码进行必要的更改以尝试一下。但是，我不明白最后两种方法table_answer 类。Model_question 和 Model_answer 方法什么都不做？ 这是独立的 java 类而不是方法 哦，现在我明白了。因为我已经有了问题和答案对象，它使工作变得更容易了。我肯定会的。这是我经过几天的搜索后得到的最有用的答案！由于我现在知道如何插入，当我想从 sqlite 检索问题以显示在 textView 中时，我是否使用类似 - model.getQuestionDescription() 的东西？ 嗨@dipali。我还没有。我在从 sqlite 检索值时遇到了一些挑战。在你向我展示了如何保存它们之后。我正在从 mysql 下载测验server.then 在一个活动中传递然后将它们保存到 sqlite。我有一个 FragmentStatePagerAdapter 类，然后将每个问题传递到一个片段上。所以我被困在寻呼机适配器类以及如何从 sqlite db 中检索值。【参考方案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.htmlHow 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）
【讨论】：
感谢 @user2450263 的建议。我正在研究 for each 循环，特别是因为很明显我需要在这里。我的代码中已经有了问题和答案对象。所以我我会执行你和 dipali 的建议，让你们知道它是如何工作的。【参考方案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;
【讨论】：
@dipali.i 正在进行必要的更改，并会在几分钟后通知您。 在对您上面显示的内容进行更改后，我不断收到我在编辑中添加的错误异常。 好吧@dipali.tomorrow 还是可以的。非常感谢你和我一起浏览这段代码。非常有耐心！ :-) 不过要澄清一下，我是说在你的第二个回答之后，我对如何检索数据进行了这些更改，但现在我得到一个“无法打开数据库错误异常”和“表答案类的空指针异常” .明天见:-) @naffie你有没有打开sqldb？ @diplai 我有。在所有必要的地方。让我在我的代码中告诉你。我会添加另一个编辑。我会尽量坚持相关部分，因为它很长跨度>以上是关于遍历嵌套的json对象并将数据保存在android sqlite的主要内容，如果未能解决你的问题，请参考以下文章