Question

我现在已经建立数据库我想向数据库添加一些数据如何通过code.anyone来帮助我存储我的数据库的价值。提前致谢。我尝试插入但未能这样做

这是我的dbhelper类。

    public class FoodDbHelper extends SQLiteOpenHelper {
private static final String DATABASE_NAME = "PKFM.DB";
private static final int DATABASE_VERSION = 1;
private static final String CREATE_QUERY =
        "CREATE TABLE "+ Food.NewDishInfo.TABLE_NAME+"("
                + Food.NewDishInfo.DISH_NAME+" TEXT NOT NULL,"
                + Food.NewDishInfo.DISH_QUANTITY+" TEXT NOT NULL,"
                + Food.NewDishInfo.DISH_CALORIE+" INTEGER,"
                + Food.NewDishInfo.DISH_FAT+" TEXT NOT NULL,"
                + Food.NewDishInfo.DISH_PROTEIN+" TEXT NOT NULL,"
                + Food.NewDishInfo.DISH_SUGAR+" TEXT NOT NULL,"
                + Food.NewDishInfo.DISH_CARBOHYDRATES+" TEXT NOT NULL);";
public FoodDbHelper(Context context)
{
    super(context,DATABASE_NAME,null,DATABASE_VERSION);
    Log.e("DATABASE OPERATION","Database created / opened...");
}
@Override
public void onCreate(SQLiteDatabase db) {
    db.execSQL(CREATE_QUERY);
    Log.e("DATABASE OPERATION","Table created...");


}
public void addInformations(String name ,String quantity, Integer calorie, String fat ,
                            String protein,String sugar,String carbohydrates, SQLiteDatabase db){


    ContentValues contentValues = new ContentValues();
    contentValues.put(Food.NewDishInfo.DISH_NAME,name);
    contentValues.put(Food.NewDishInfo.DISH_QUANTITY,quantity);
    contentValues.put(Food.NewDishInfo.DISH_CALORIE,calorie);
    contentValues.put(Food.NewDishInfo.DISH_FAT,fat);
    contentValues.put(Food.NewDishInfo.DISH_PROTEIN,protein);
    contentValues.put(Food.NewDishInfo.DISH_SUGAR,sugar);
    contentValues.put(Food.NewDishInfo.DISH_CARBOHYDRATES,carbohydrates);

    db.insert(Food.NewDishInfo.TABLE_NAME, null, contentValues);


    Log.e("DATABASE OPERATION","one row inserted...");
}

我做了这样的事情，但没有插入数据

foodDbHelper = new FoodDbHelper(context);
foodDbHelper.addInformations("Paratha", "1 piece", 260,"8.99 g","5.16 g","2.18 g","38.94 g",sqLiteDatabase);

Answer 1

使用this.getWritableDatabase（）;喜欢这个功能希望它能帮到你..

//Declare variable like this

private static final String TABLE_FRIEND_MASTER = "friend_master";  

private static final String CREATE_TABLE_FRIEND_MASTER = "CREATE TABLE IF NOT EXISTS " + TABLE_FRIEND_MASTER + "(" + KEY_ID + " INTEGER PRIMARY KEY," + KEY_FRIEND_ID + " TEXT," + KEY_FRIEND_FNAME + " TEXT," + KEY_FRIEND_LNAME + " TEXT," + KEY_FRIEND_EMAIL + " TEXT," + KEY_FRIEND_MOBILE + " TEXT," + KEY_FRIEND_USERNAME + " TEXT,"
            + KEY_FRIEND_COUNTRY_CODE + " TEXT);";


@Override
    public void onCreate(SQLiteDatabase db) {
        db.execSQL(CREATE_TABLE_FRIEND_REQUEST_MASTER);
    }



    public int addFriend(String friendId, String friendFname, String friendLname, String friendEmail, String friendMobile, String friendUsername, String friendCountryCode) {
        synchronized (Lock) {
            SQLiteDatabase db = this.getWritableDatabase();
            ContentValues values = new ContentValues();
            values.put(KEY_FRIEND_ID, "" + friendId);
            values.put(KEY_FRIEND_FNAME, "" + friendFname);
            values.put(KEY_FRIEND_LNAME, "" + friendLname);
            values.put(KEY_FRIEND_EMAIL, "" + friendEmail);
            values.put(KEY_FRIEND_MOBILE, "" + friendMobile);
            values.put(KEY_FRIEND_USERNAME, "" + friendUsername);
            values.put(KEY_FRIEND_COUNTRY_CODE, "" + friendCountryCode);
            // insert row
            int comment_row_id = (int) db.insert(TABLE_FRIEND_MASTER, null, values);
            return comment_row_id;
        }
    }

Android将数据插入数据库问题

1 个答案: