Question

数据库本身就是用自己的食谱（食谱）和食材（RecipeIngredient）储存鸡尾酒。每个用户（用户）都有自己的“食品室”（UserIngredients），在这里他们可以存储他们在家里的食材。此查询现在应该向他们展示他们可以混合的鸡尾酒

我有以下查询：

SELECT u.User_Name, 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 = 'DgRkQztkvUhotfSf53l7ciiI8rOhKtuvoPqCTvdlBXWTn9cYxz'

并且想知道是否可以为每个配方获得一个“r.Recipe_Name”而不是每个配料一个。

我的tablelayout如下：

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_Nickname TEXT,
    User_EMail TEXT,
    User_Session VARCHAR(50) UNIQUE,
    User_Admin BOOLEAN
);
CREATE TABLE Recipe
(
    Recipe_ID INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
    Recipe_Name TEXT NOT NULL,
    Recipe_Clicks INT,
    Recipe_Description TEXT
);
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)
);

Answer 1

要使用此数据库获得所需结果，请尝试

SELECT DISTINCT u.User_Name, 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 = 'DgRkQztkvUhotfSf53l7ciiI8rOhKtuvoPqCTvdlBXWTn9cYxz'

我的猜测是用户创建了Recipies，为什么不将User_ID添加到Receipe？

Answer 2

你可以使用不同的

SELECT DISTINCT u.User_Name, 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 = 'DgRkQztkvUhotfSf53l7ciiI8rOhKtuvoPqCTvdlBXWTn9cYxz'

Answer 3

尝试

SELECT u.User_Name, MAX(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 = 'DgRkQztkvUhotfSf53l7ciiI8rOhKtuvoPqCTvdlBXWTn9cYxz'
GROUP BY u.User_Name, r.Recipe_Name

不确定这一点，但听起来多个成分将具有相同的配方，因此只需选择最大值，这将返回唯一的配方名称，如果按用户名+配方名称分组，它可能会为您提供所需。

如何将结果限制为不同的结果？

3 个答案: