Код: Выделить всё
SELECT DISTINCT
r.Recipe_Name
FROM
USER u
INNER JOIN
UserIngredient ui
ON
u.User_ID = ui.User_ID
INNER JOIN
RecipeIngredient ri
ON
ui.Ingredient_ID = ri.Ingredient_ID
INNER JOIN
Ingredient i
ON
ri.Ingredient_ID = i.Ingredient_ID
INNER JOIN
Recipe r
ON
ri.Recipe_ID = r.Recipe_ID
WHERE
u.User_Session = SessionID
Код: Выделить всё
CREATE TABLE User
(
User_ID INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
User_Pass TEXT NOT NULL,
User_Name TEXT NOT NULL,
User_Surname TEXT NOT NULL,
User_EMail VARCHAR(255) UNIQUE,
User_Session VARCHAR(50) UNIQUE,
User_Admin BOOLEAN,
User_Newsletter BOOLEAN
);
CREATE TABLE Recipe
(
Recipe_ID INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
Recipe_Name VARCHAR(255) UNIQUE NOT NULL,
Recipe_Clicks INT,
Recipe_Description TEXT,
Recipe_Image VARCHAR(255) DEFAULT 'https://placehold.it/400x500'
);
CREATE TABLE Ingredient
(
Ingredient_ID INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
Ingredient_Name VARCHAR(255) UNIQUE,
Ingredient_Description TEXT
);
CREATE TABLE RecipeIngredient
(
RecipeIngredient_ID INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
RecipeIngredient_Amount DECIMAL(8,2) NOT NULL,
MeasuringUnit_ID INT NOT NULL,
Recipe_ID INT NOT NULL,
Ingredient_ID INT NOT NULL,
FOREIGN KEY (MeasuringUnit_ID) REFERENCES MeasuringUnit(MeasuringUnit_ID),
FOREIGN KEY (Recipe_ID) REFERENCES Recipe(Recipe_ID),
FOREIGN KEY (Ingredient_ID) REFERENCES Ingredient(Ingredient_ID)
);
CREATE TABLE UserIngredient
(
UserIngredient_ID INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
Ingredient_ID INT NOT NULL,
User_ID INT NOT NULL,
FOREIGN KEY(Ingredient_ID) REFERENCES Ingredient(Ingredient_ID),
FOREIGN KEY(User_ID) REFERENCES User(User_ID)
);
Я получаю каждый рецепт, в котором есть один ингредиент.
Результат, который я получаю. хочу:
Все рецепты, в которых у меня есть ВСЕ ингредиенты.
Подробнее здесь: https://stackoverflow.com/questions/368 ... dients-for
Мобильная версия