我在本地服务器上有一个MySQL数据库。我使用PHP和JSON通过HTTP从MySQL数据库中获取数据。在Android应用程序中,使用JSON结果填充SQLite数据库。我已经验证了SQLite数据库实际上是通过运行类似于
的测试查询来填充的Cursor cursor = database.rawQuery("SELECT recipe_id, ingredient_id, name FROM mm_ingredient_in_recipe_groups", null);
然后迭代生成的光标以查看数据是否存在。存在的数据存储在三个表中:
mm_ingredients
,含有食谱成分。mm_recipes
,包含食谱数据。mm_ingredient_in_recipe
,其中包含用于加入mm_recipes
和mm_ingredients
中数据的数据(即解决多对多关系)。我想要做的是运行一个查询,为参数配方名称选择成分。这意味着我必须与mm_ingredients
以及mm_ingredient_in_recipe
一起加入mm_recipes
。
我尝试过运行以下查询:
database.rawQuery("SELECT ingredient.name, ingredient.base_unit FROM mm_ingredients AS ingredient INNER JOIN mm_ingredient_in_recipe AS ir ON ir.ingredient_id = ingredient._id INNER JOIN mm_recipes AS recipe ON ir.recipe_id = recipe._id WHERE recipe.name = ?", new String[] {"Spicy tomatsuppe"});
返回一个空光标。但是,在我的MySQL数据库中,运行相应的查询
SELECT ingredient.name, ingredient.base_unit FROM mm_ingredients AS ingredient INNER JOIN mm_ingredient_in_recipe AS ir ON ir.ingredient_id = ingredient.id INNER JOIN mm_recipes AS recipe ON ir.recipe_id = recipe.id WHERE recipe.name = "Spicy tomatsuppe";
返回预期的数据。我在这做错了什么?似乎我运行的每个查询都可以正常工作,直到我开始包含INNER JOIN。 (在SQLite中,id字段被命名为_id,据我所知应该是最佳实践。)
编辑:添加用于写入数据库的代码:
下面是我如何填充表mm_recipes
的数据库的示例。它与其他两个表类似。我将JSONArray.optString()
用于SQLite的TEXT
数据类型,JSONArray.optInt()
用于INTEGER
,JSONArray.optDouble()
用于REAL
。
public static final String TABLE_NAME = "mm_recipes";
public static final String TABLE_NAME_IN_JSON = "mm_recipes";
public static final String COLUMN_ID = "_id";
public static final String COLUMN_NAME = "name";
public static final String COLUMN_IN_APP = "in_app";
public static final String COLUMN_DIFFICULTY = "difficulty";
public static final String COLUMN_TIME_CONSUMPTION_UPPER = "time_consumption_upper";
public static final String COLUMN_TIME_CONSUMPTION_LOWER = "time_consumption_lower";
public static final String COLUMN_PREPARATION_TIME = "preparation_time";
public static final String COLUMN_PORTIONS = "portions";
public static final String COLUMN_PORTIONS_UNIT = "portions_unit";
public static final String COLUMN_DESCRIPTION = "description";
public static final String COLUMN_TIPS = "tips";
public static final String COLUMN_USE_INGREDIENT_GROUPS = "use_ingredient_groups";
public static final void populateDatabase(SQLiteDatabase database, JSONObject databaseData) throws JSONException{
JSONArray table_data = databaseData.optJSONArray(TABLE_NAME_IN_JSON);
ContentValues contentValues = new ContentValues();
for (int i = 0; i < table_data.length(); i++){
JSONObject data = table_data.getJSONObject(i);
contentValues.clear();
contentValues.put(COLUMN_NAME, data.optString(COLUMN_NAME));
contentValues.put(COLUMN_IN_APP, data.optInt(COLUMN_IN_APP));
contentValues.put(COLUMN_DIFFICULTY, data.optString(COLUMN_DIFFICULTY));
contentValues.put(COLUMN_TIME_CONSUMPTION_UPPER, data.optInt(COLUMN_TIME_CONSUMPTION_UPPER));
contentValues.put(COLUMN_TIME_CONSUMPTION_LOWER, data.optInt(COLUMN_TIME_CONSUMPTION_LOWER));
contentValues.put(COLUMN_PREPARATION_TIME, data.optInt(COLUMN_PREPARATION_TIME));
contentValues.put(COLUMN_PORTIONS, data.optInt(COLUMN_PORTIONS));
contentValues.put(COLUMN_PORTIONS_UNIT, data.optString(COLUMN_PORTIONS_UNIT));
contentValues.put(COLUMN_DESCRIPTION, data.optString(COLUMN_DESCRIPTION));
contentValues.put(COLUMN_TIPS, data.optString(COLUMN_TIPS));
contentValues.put(COLUMN_USE_INGREDIENT_GROUPS, data.optInt(COLUMN_USE_INGREDIENT_GROUPS));
database.insert(TABLE_NAME, null, contentValues);
}
}
作为参考,这里也是一个如何创建mm_recipe
表的示例。在我的班级mm_recipe
:
public static final String DATATYPE_ID = "INTEGER PRIMARY KEY AUTOINCREMENT NOT NULL";
public static final String DATATYPE_NAME = "TEXT NOT NULL";
public static final String DATATYPE_IN_APP = "INTEGER DEFAULT 0";
public static final String DATATYPE_DIFFICULTY = "TEXT NOT NULL";
public static final String DATATYPE_TIME_CONSUMPTION_UPPER = "INTEGER NOT NULL";
public static final String DATATYPE_TIME_CONSUMPTION_LOWER = "INTEGER NOT NULL";
public static final String DATATYPE_PREPARATION_TIME = "INTEGER";
public static final String DATATYPE_PORTIONS = "INTEGER NOT NULL";
public static final String DATATYPE_PORTIONS_UNIT = "TEXT NOT NULL";
public static final String DATATYPE_DESCRIPTION = "TEXT";
public static final String DATATYPE_TIPS = "TEXT";
public static final String DATATYPE_USE_INGREDIENT_GROUPS = "INT";
/**Create table statement.*/
public static final String CREATE_TABLE =
"CREATE TABLE" + " " + TABLE_NAME + " (" +
COLUMN_ID + " " + DATATYPE_ID + ", " +
COLUMN_NAME + " " + DATATYPE_NAME + ", " +
COLUMN_IN_APP + " " + DATATYPE_IN_APP + ", " +
COLUMN_DIFFICULTY + " " + DATATYPE_DIFFICULTY + ", " +
COLUMN_TIME_CONSUMPTION_UPPER + " " + DATATYPE_TIME_CONSUMPTION_UPPER + ", " +
COLUMN_TIME_CONSUMPTION_LOWER + " " + DATATYPE_TIME_CONSUMPTION_LOWER + ", " +
COLUMN_PREPARATION_TIME + " " + DATATYPE_PREPARATION_TIME + ", " +
COLUMN_PORTIONS + " " + DATATYPE_PORTIONS + ", " +
COLUMN_PORTIONS_UNIT + " " + DATATYPE_PORTIONS_UNIT + ", " +
COLUMN_DESCRIPTION + " " + DATATYPE_DESCRIPTION + ", " +
COLUMN_TIPS + " " + DATATYPE_TIPS + ", " +
COLUMN_USE_INGREDIENT_GROUPS + " " + DATATYPE_USE_INGREDIENT_GROUPS + ");";
在我的SQLiteOpenHelper
课程中:
@Override
public void onCreate(SQLiteDatabase database) {
database.execSQL(Table_mm_recipes.CREATE_TABLE);
database.execSQL(Table_mm_ingredient_in_recipe.CREATE_TABLE);
database.execSQL(Table_mm_ingredients.CREATE_TABLE);
}
答案 0 :(得分:0)
事实证明这是(正如预期的)一些愚蠢的错误。我会把这个留在这里以防其他人犯同样的错误。
因为我每次都设置了代码来删除创建数据库中的所有内容,所以这样:
database.execSQL("DELETE FROM " + Table_mm_recipes.TABLE_NAME);
database.execSQL("DELETE FROM " + Table_mm_ingredient_in_recipe.TABLE_NAME);
database.execSQL("DELETE FROM " + Table_mm_ingredients.TABLE_NAME);
某种方式,数据发生了不好的事情。这样做没有多大意义,因为你可以创建数据库并在SQLiteOpenHelper.onCreate()
插入数据并且没问题。如果您需要在设置任何版本控制之前更新数据,只需卸载并重新安装应用程序即可。然后,如果愿意,您可以在SQLiteOpenHelper.onUpgrade()
销毁数据。我不确定如何数据搞砸了,但他们肯定是。
的奇怪之处在于,我仍然可以通过做一个简单的
看到数据库中有数据Cursor cursor = database.rawQuery("SELECT recipe_id, ingredient_id, name FROM mm_ingredient_in_recipe_groups", null);
并从之前迭代该游标。我在INNER JOIN
语句之前直接对所有表执行了此操作,但它们仍然无效。它可能与错误删除/创建数据库有关。
除此之外,上述所有代码都应该有效且有效。