Question

食物可以是简单的或Agregate ..如果它的简单它有一种类型..如果agregate它有一个简单的食物组成我需要创建一个查询来选择所有出售所有食品类型=肉类的agregate菜肴的餐馆。我尝试了很多东西，但我无法解决问题。

    create table Restaurant(
        nameR VARCHAR(80),
        primary key (nameR));

    create table Sold(
        nameF VARCHAR(80),
        nameR VARCHAR(80),
        date VARCHAR(20);
        constraint c3 primary key(nameF,nameR,date),
        foreign key(nameF) references Dish(nameF),
        foreign key(nameR) references Restaurante(nameR),
        foreign key(date) references Data(date));

    create table Dishe(
        nameF VARCHAR(80),
        primary key (nameF),
        foreign key (nameF) references Food(nameF));

   create table Simple( 
       nameF VARCHAR(80), 
      type VARCHAR(80),
      primary key (nameF), 
       foreign key (nameF) references Food(nameF));

    create table Agregate(
       nameF VARCHAR(80),
       calorias double,
       primary key (nameF),
       foreign key (nameF) references Food(nameF));

    create table Composition(
       nameAgg VARCHAR(80),
        nameS VARCHAR(80),
        quantidade integer,
       constraint c1 primary key(nameAgg,nameS),
       foreign key(nameAgg) references Agregate(nameF),
      foreign key(nameS) references Simple(nameF));

    create table Food( 
         nameF varchar(80)
         primary key (nameF));

我认为我应该使用分组吗？告诉我你的想法

感谢

Answer 1

你的SQL有一些语法错误，它还引用外键“已售出”中的数据表，我不知道为什么“Dishe”表存在 - 它似乎只存储与食物相同的数据？

无论如何，这是用于创建表格的更正SQL：

create table Restaurant(
    nameR VARCHAR(80),
    primary key (nameR)
);

create table Food( 
    nameF varchar(80), 
    type varchar(80),
    primary key (nameF)
);

create table Dishe(
    nameF VARCHAR(80),
    primary key (nameF),
    foreign key (nameF) references Food(nameF)
);

create table Sold(
    nameF VARCHAR(80),
    nameR VARCHAR(80),
    date VARCHAR(20),
    constraint c3 primary key(nameF,nameR,date),
    foreign key(nameF) references Dishe(nameF),
    foreign key(nameR) references Restaurant(nameR)
);

我们可以插入一些测试数据：

INSERT INTO Restaurant (nameR) VALUES ('McDonalds');
INSERT INTO Restaurant (nameR) VALUES ('Vital Ingredient');
INSERT INTO Restaurant (nameR) VALUES ('David Food');
INSERT INTO Food (nameF, type) VALUES ('Burger', 'meat');
INSERT INTO Food (nameF, type) VALUES ('Salad', 'veg');
INSERT INTO Dishe (nameF) VALUES ('Burger');
INSERT INTO Dishe (nameF) VALUES ('Salad');
INSERT INTO Sold VALUES ('Salad', 'David Food', '2013-12-05');
INSERT INTO Sold VALUES ('Burger', 'McDonalds', '2013-12-04');
INSERT INTO Sold VALUES ('Burger', 'McDonalds', '2013-12-05');
INSERT INTO Sold VALUES ('Burger', 'Vital Ingredient', '2013-12-05');
INSERT INTO Sold VALUES ('Salad', 'Vital Ingredient', '2013-12-04');

这创造了三家餐厅，只有一家餐厅只售卖了肉类菜肴。

我能想到的两种方法可以归还售卖所有肉类菜肴的餐厅。第一个不使用分区 - 它使用GROUP_CONCAT功能将每个餐厅提供的所有类型的菜肴粘在一起，看看这是否等于“肉”：

SELECT 
    Sold.nameR
FROM
    Sold
INNER JOIN
    Food
ON
    Food.nameF = Sold.nameF
GROUP BY 
    Sold.nameR
HAVING
    GROUP_CONCAT(DISTINCT Food.type) = 'meat'
;

第二个使用分区来计算每个餐馆出售的肉类菜肴的数量，并将其除以服务的菜肴总数，并检查它是否等于1 - 即100％的菜肴是肉。由于子查询，此方法可能比上面的方法慢：

SELECT 
    Restaurant.nameR
FROM
    Restaurant
WHERE
    (
    SELECT 
        COUNT(*) 
    FROM 
        Sold 
    INNER JOIN 
        Food 
    ON 
        Food.nameF = Sold.nameF 
    WHERE 
        Food.type = 'meat' 
    AND 
        Sold.nameR = Restaurant.nameR 
    ) / (
    SELECT 
        COUNT(*) 
    FROM 
        Sold 
    INNER JOIN 
        Food 
    ON 
        Food.nameF = Sold.nameF 
    WHERE 
        Sold.nameR = Restaurant.nameR 
    ) = 1
;

使用division来选择mysql中的所有元素

1 个答案: