Question

在我的应用程序中，我希望将数据保存在数据库中。

这是我的SQLiteHelper代码

public class UserSqliteHelper extends SQLiteOpenHelper {
    private final String LOGCAT = "JBF/SQLite";
    private static final int DATABASE_VERSION = 1;

    // Database Name
    private static final String DATABASE_NAME = "jbfjsonEntityDB";

    private static final String TABLE_NAME = "jbfjsonEntity";

    private static final String KEY_JSON = "json";
    private static final String KEY_URL_PATH = "url_path";
    private static final String KEY_TIME = "added_on";

    public UserSqliteHelper(Context context) {
        super(context, "dictionarysqlitehelper.db", null, 1);
        Log.d(LOGCAT, "Created");
    }

    @Override
    public void onCreate(SQLiteDatabase db) {
        String CREATE_CONTACTS_TABLE = "CREATE TABLE " + TABLE_NAME + "("
                + KEY_JSON + " TEXT, "
                + KEY_TIME + " TIMESTAMP NOT NULL DEFAULT current_timestamp, "
                + KEY_URL_PATH  + " TEXT )";
        db.execSQL(CREATE_CONTACTS_TABLE);
    }

    @Override
    public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
        String query = "DROP TABLE IF EXISTS " + TABLE_NAME ;
        db.execSQL(query); onCreate(db);
    }

    public void addJsonEntity(JsonEntity jsonEntity) {
        SQLiteDatabase db = this.getWritableDatabase();

        ContentValues values = new ContentValues();
        values.put(KEY_JSON, jsonEntity.getJson());
        values.put(KEY_URL_PATH, jsonEntity.getUrl_path());

        // Inserting Row
        db.insert(TABLE_NAME, null, values);
        db.close();
    }

    public JSONObject getJsonByUrl(String url) {
        String json = "";
        SQLiteDatabase db = this.getReadableDatabase();

        try {
//            Cursor c = db.query(TABLE_NAME, null, KEY_URL_PATH + "=?", new String[]{url}, null, null, null);
            String selectQuery = "SELECT * FROM " + TABLE_NAME + " where " + KEY_URL_PATH + "='"+url+"'";
            Cursor c = db.rawQuery(selectQuery, null);

            if (c == null) {
                return null;
            } else {
                c.moveToFirst();
                json =c.getString(c.getColumnIndex(KEY_JSON));
                if (json != null) {
                    return new JSONObject(json);
                } else {
                    return null;
                }
            }
        } catch (Exception e) {
            e.printStackTrace();
        }

        return null;
    }
}

当我从我的活动中打电话时

UserSqliteHelper sqliteHelper = new UserSqliteHelper(SplashActivity.this);
sqliteHelper.getWritableDatabase();
sqliteHelper.addJsonEntity(new JsonEntity(STRING_CONFIGS_URL,response.toString()));
System.out.println("json  ==== "+sqliteHelper.getJsonByUrl(GET_USER_INFO_URL));

我总是遇到这个错误

android.database.CursorIndexOutOfBoundsException: Index 0 requested, with a size of 0

有谁能告诉我这里做错了什么。为什么我无法获取数据库值？

Answer 1

查询没有匹配任何数据。 moveToFirst()失败，光标未指向有效行。你应该检查moveToFirst()是否成功 - 它返回一个布尔值。

为什么它与任何数据都不匹配是因为您通过不同的密钥存储和检索数据：STRING_CONFIGS_URL和GET_USER_INFO_URL。

Answer 2

而不是c == null尝试c.getColumnCount == 0

无法从sqlite表中获取值

2 个答案: