有人可以建议一种方法来修改JSONParser的getQuestionJSONFromUrl（）方法，以便将每个问题存储为Android中自己的对象吗？通过PHP JSON读取多个结果从一个在浏览器中看起来像这样的SQL表返回：
{"category":"elections","id":"0","title":"Who will you vote for in November's Presidential election?","published":"2012-04-02","enddate":"2012-04-30","responsetype":"0"}
{"category":"elections","id":"2","title":"Question title, ladies and gents","published":"2012-04-02","enddate":"2012-04-30","responsetype":"1"}
目前，结果甚至不包括端支架和起始支架之间的空间。但我可以添加到我的PHP：echo“\ n”;当JSON读出时，这会在两行之间给我一个空格。所以现在显然有两排填料含量。最终SQL表中会有真正的内容。
我希望能够将这些行分解为对象（可能是我想要的本地SQLite数据库？），这样我就可以使用它们在片段中将每个行显示为屏幕上的一个tablerow。我并不太担心屏幕方面，但是将数据变为可行的形式是一个问题。目前，我的代码只将第一组大括号存储为JSON对象。以下是所有相关代码：
public UserFunctions(){
jsonParser = new JSONParser();
}
public JSONObject getQuestions(String category) {
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("tag", question_tag));
params.add(new BasicNameValuePair("category", category));
JSONObject json = jsonParser.getQuestionJSONFromUrl(questionURL, params);
return json;
}
public class JSONParser {
static InputStream is = null;
static JSONObject jObj = null;
static JSONObject[] jsonArray = null;
static String json = "";
// constructor
public JSONParser() {
}
public JSONObject getQuestionJSONFromUrl(String url, List<NameValuePair> params) {
// Making HTTP request
try {
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
httpPost.setEntity(new UrlEncodedFormEntity(params));
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
Log.v("while", line);
sb.append(line + "\n");
//Log.v("err", line);
}
is.close();
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
}
有人可以建议一种方法来修改JSONParser的getQuestionJSONFromUrl（）方法，以便将每个问题存储为Android中自己的对象吗？我有一个本地SQLite DB，我可以在其中添加一个或两个方法来添加第二个表等：
package library;
import java.util.HashMap;
import android.content.ContentValues;
import android.content.Context;
import android.database.Cursor;
import android.database.sqlite.SQLiteDatabase;
import android.database.sqlite.SQLiteOpenHelper;
public class DatabaseHandler extends SQLiteOpenHelper {
// All Static variables
// Database Version
private static final int DATABASE_VERSION = 1;
// Database Name
private static final String DATABASE_NAME = "android_api";
// Login table name
private static final String TABLE_LOGIN = "login";
// Login Table Columns names
private static final String KEY_ID = "id";
private static final String KEY_NAME = "name";
public static final String KEY_EMAIL = "email";
private static final String KEY_UID = "uid";
private static final String KEY_CREATED_AT = "created_at";
public DatabaseHandler(Context context) {
super(context, DATABASE_NAME, null, DATABASE_VERSION);
}
// Creating Tables
@Override
public void onCreate(SQLiteDatabase db) {
String CREATE_LOGIN_TABLE = "CREATE TABLE " + TABLE_LOGIN + "("
+ KEY_ID + " INTEGER PRIMARY KEY,"
+ KEY_NAME + " TEXT,"
+ KEY_EMAIL + " TEXT UNIQUE,"
+ KEY_UID + " TEXT,"
+ KEY_CREATED_AT + " TEXT" + ")";
db.execSQL(CREATE_LOGIN_TABLE);
}
// Upgrading database
@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
// Drop older table if existed
db.execSQL("DROP TABLE IF EXISTS " + TABLE_LOGIN);
// Create tables again
onCreate(db);
}
/**
* Storing user details in database
* */
public void addUser(String name, String email, String uid, String created_at) {
SQLiteDatabase db = this.getWritableDatabase();
ContentValues values = new ContentValues();
values.put(KEY_NAME, name); // Name
values.put(KEY_EMAIL, email); // Email
values.put(KEY_UID, uid); // Email
values.put(KEY_CREATED_AT, created_at); // Created At
// Inserting Row
db.insert(TABLE_LOGIN, null, values);
db.close(); // Closing database connection
}
/**
* Getting user data from database
* */
public HashMap<String, String> getUserDetails(){
HashMap<String,String> user = new HashMap<String,String>();
String selectQuery = "SELECT * FROM " + TABLE_LOGIN;
SQLiteDatabase db = this.getReadableDatabase();
Cursor cursor = db.rawQuery(selectQuery, null);
// Move to first row
cursor.moveToFirst();
if(cursor.getCount() > 0){
user.put("name", cursor.getString(1));
user.put("email", cursor.getString(2));
user.put("uid", cursor.getString(3));
user.put("created_at", cursor.getString(4));
}
cursor.close();
db.close();
// return user
return user;
}
/**
* Getting user login status
* return true if rows are there in table
* */
public int getRowCount() {
String countQuery = "SELECT * FROM " + TABLE_LOGIN;
SQLiteDatabase db = this.getReadableDatabase();
Cursor cursor = db.rawQuery(countQuery, null);
int rowCount = cursor.getCount();
db.close();
cursor.close();
// return row count
return rowCount;
}
/**
* Re crate database
* Delete all tables and create them again
* */
public void resetTables(){
SQLiteDatabase db = this.getWritableDatabase();
// Delete All Rows
db.delete(TABLE_LOGIN, null, null);
db.close();
}
}
答案 0 :(得分：1)
make getQuestionJSONFromUrl返回字符串。然后使用该字符串创建JSONObject并解析它。请看下面的例子。
JSONObject jsonobj=new JSONObject(str);
String category=jsonobj.getString("category");
String title=jsonobj.getString("title");
int id=jsonobj.getInt("id");
String published=jsonobj.getString("published");
String enddate=jsonobj.getString("enddate");
int responsetype=jsonobj.getInt("responsetype");
System.out.println(category+" "+title +" "+id +" "+published +" "+enddate +" "+responsetype);
答案 1 :(得分：1)
如果你的回复内容更多，那么使用for循环来检索所有数据，只需按照下面的代码，希望它对你有用。
JSONArray Arraylist = new JSONArray();
Arraylist=new JSONObject(output).getJSONObject("category").getJSONArray("title");
for (int i = 0; i < Arraylist.length(); i++)
{
JSONObject headObject = Arraylist.getJSONObject(i);
String Question_title= headObject.optString("Question title");
String published=headObject.optString("published");
String enddate=headObject.optString("enddate");
}
答案 2 :(得分：0)
这是使PHP脚本返回正确编码的JSON数据的解决方案。我目前仍在研究如何让Android解析为JSONArray：https://stackoverflow.com/a/10019165/1231943