php - 存储 JSON 的多个结果

Question

任何人都可以建议一种方法来修改 JSONParser 的 getQuestionJSONFromUrl() 方法，以便它将每个问题存储为 Android 中自己的对象吗？通过 PHP JSON 返回的 SQL 表读取多个结果，在浏览器中如下所示:

{"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，结果甚至不包括结束大括号和开始大括号之间的空格。但我可以添加到我的 php: echo "\n";当 JSON 读出时，这会在两行之间留出一个空格。所以现在显然有两行填充内容。最终该 SQL 表中将包含真实内容。

我希望能够将这些行分解为对象(我想可能是我的本地 SQLite 数据库？)，以便我可以使用它们在屏幕上的 fragment 中将每一行显示为表格行。我不太担心屏幕方面，但将数据转化为可用的形式是一个问题。目前，我的代码仅将第一组大括号存储为 JSON 对象。这是所有相关代码:

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;

    }

任何人都可以建议一种方法来修改 JSONParser 的 getQuestionJSONFromUrl() 方法，以便它将每个问题存储为 Android 中自己的对象吗？我确实有一个本地 SQLite 数据库，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();
    }

}

Answer 1

使getQuestionJSONFromUrl返回字符串。然后使用该字符串创建 JSONObject 并解析它。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);