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多個結果，任何人都可以提出一個方法）來修改JSONParser的getQuestionJSONFromUrl（方法，這樣它存儲每一個問題作爲自己的Android對象？通過從SQL表看起來像這樣在瀏覽器PHP JSON返回讀取多個結果：存儲從JSON
{"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，這樣做的結果還不包括年底支柱之間的空間，並開始梅開二度。但我可以添加到我的PHP：回聲「\ n」;當JSON讀出時，這會給我兩行之間的空間。所以現在顯然是兩排填充物。最終會在該SQL表中存在真實的內容。
我希望能夠將這些行分解成對象（可能是我本地的SQLite數據庫我想？），以便我可以使用它們在屏幕上顯示每個片段作爲片段。我並不太擔心屏幕方面的問題，但將數據轉換爲可行的形式是一個問題。目前，我的代碼只將第一組大括號存儲爲JSON對象。這裏的所有相關代碼：
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;
}
任何人都可以提出一個方法）來修改JSONParser的getQuestionJSONFromUrl（方法，這樣它存儲每一個問題作爲自己的Android對象？我有一個本地SQLite數據庫中，我可以添加一個或兩個的方法來添加第二個表等：
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();
}
}
這是非常有用的，並且可以像廣告一樣工作 - 但是如果方法getQuestion ...只運行一次，並且有20個（任意數字，有時會更多，有時它會更少）得到返回的問題，請問該方法需要改變，以適應從SQL 20個分開行？ – Davek804 2012-04-03 04:04:57
使用for循環，用於檢索一個以上的時間，如對於 （INT I = 0; I
wolverine
2012-04-03 04:16:50
我花了一整天在這方面的工作，我仍然不能得到它的工作。我已經能夠做的最好的是檢測「} {」（二JSON結束/開始），並放置在一個字符串的第一部分，並刪除了下半場。我發現我需要的方法不在那裏。 – Davek804 2012-04-04 03:58:43