更新了表,但没有返回正确的值

时间:2018-04-21 14:57:21

标签: java android sql sqlite

可能是一件简单的事情,但我花了很长时间才弄明白。

我想保存评级栏评级,然后在重新打开活动时加载该评级。但它总是返回0.

我有一个带有评级栏的食谱表:

final String CREATE_TABLE_RECIPES = "CREATE TABLE IF NOT EXISTS " +
        TABLE_RECIPES + "(" +
        RECIPE_ID + " INTEGER PRIMARY KEY AUTOINCREMENT, " +
        RECIPE_NAME + " TEXT, " +
        RECIPE_INSTRUCTIONS + " TEXT, " +
        RECIPE_RATING + " FLOAT " +
        ")";

我有一个onRatingBarChange侦听器将评级保存到表:

(id和评级的值是正确的)

ratingbar.setOnRatingBarChangeListener(new RatingBar.OnRatingBarChangeListener() {
        @Override
        public void onRatingChanged(RatingBar ratingBar, float rating, boolean fromUser) {
            db.addRating(id, rating);
       }
});

db class中的addRating方法:

public void addRating(int id, float rating) {
    String SET_RATING =
            "UPDATE " + TABLE_RECIPES +
                    " SET " + RECIPE_RATING + " = " + rating +
                    " WHERE " + RECIPE_ID + " = " + id;

    SQLiteDatabase db = this.getWritableDatabase();
    db.rawQuery(SET_RATING, null);
    db.close();
}

最后,db class中的getRating方法:

public float getRating(int id) {
    String GET_RATING =
            "SELECT " + RECIPE_RATING +
                    " FROM " + TABLE_RECIPES +
                    " WHERE " + RECIPE_ID + " = "  + id;
    SQLiteDatabase db = this.getReadableDatabase();
    Cursor c = db.rawQuery(GET_RATING, null);
    c.moveToFirst();
    float rating =  c.getFloat(0);
    Log.i("getRating", "gotRating: " + rating);
    return rating;
}

返回的评级始终为0,因此要么没有正确保存,要么没有正确获得该值。

1 个答案:

答案 0 :(得分:1)

db.rawQuery(SET_RATING, null);

您应该execSQL()使用UPDATE查询,而不是rawQuery()。 (或其中一个便利包装,如update()。)

rawQuery()编译底层SQL但不执行它; execSQL()编译并运行SQL。