带有db的Android问题在DataBase中保存相同名称和NULL值

时间:2015-06-06 10:49:21

标签: android mysql

我创建了一个数据库,它保存了相同的名称,并且在db.i中也是null值,当我输入名称和其他值时,如果名称已经存在于数据库中,则显示Toast并且在Toast中显示该名称已经存在并且执行不保存值。所以当所有值为空时我按保存它显示吐司请插入信息。任何人都帮我解决这个问题。没有回复

</MultiTrigger.Conditions>
                                <MultiTrigger.EnterActions>
                                    <BeginStoryboard >
                                        <Storyboard >
                                            <DoubleAnimation Storyboard.TargetProperty="Opacity"
                                 From="0.7" To="0.2" Duration="0:0:0.3" BeginTime="0:0:5" />
                                        </Storyboard>
                                    </BeginStoryboard>
                                </MultiTrigger.EnterActions>
                            </MultiTrigger>

// 这是我的数据插入活动

    < public class FoodDbHelper extends SQLiteOpenHelper {
    private static final String DATABASE_NAME = "MY_Calorie_DISH.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 NOT NULL,"
                    + Food.NewDishInfo.DISH_FAT+" TEXT NOT NULL,"
                    + Food.NewDishInfo.DISH_PROTEIN+" TEXT NOT NULL,"
                    + Food.NewDishInfo.DISH_SUGAR+" TEXT NOT NULL,"
                    + Food.NewDishInfo.DISH_VITAMINS+" 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 vitamins, 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_VITAMINS,vitamins);
        db.insert(Food.NewDishInfo.TABLE_NAME, null, contentValues);
        Log.e("DATABASE OPERATION","one row inserted...");
    } 
   @Override
    public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {

    }
} >

//提前谢谢

0 个答案:

没有答案