我有数据库工作正常。我使用edittext将数据插入到database.data插入是很好的工作。但与edittext如何我可以通过code.anyone来帮助我存储我的数据库的价值。 提前谢谢。
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);
答案 0 :(得分:0)
不太确定你在寻找什么...... 您还希望如何获取存储到数据库中的信息?
这似乎更像是一个用户体验问题,而不是“如何”#。 正确的方法似乎已经包含在你的代码中,无论你想要什么字符串都可以推送。