mysql - 查找具有多个链接行的行

Question

我正在做我的高中项目 我想为用户运行查询以使用多个类别搜索药物 这是 ddl，

Create table medicine (
  MedId int(11) NOT NULL AUTO_INCREMENT,
  Name varchar(30) COLLATE utf8_bin DEFAULT NULL,
  Vendor varchar(20) COLLATE utf8_bin DEFAULT NULL,
  Quantity int(11) NOT NULL,
  type text COLLATE utf8_bin NOT NULL,
  price int(11) NOT NULL,
  PRIMARY KEY (MedId)
); 

create table category(
 CatId smallint auto_increment,
 Name varchar (20),
 primary key(CatID)
);

create table MedicineCat(
MedId int,
CatId smallint,
    foreign key (MedId) references medicine(MedId),
    foreign key (CatId) references category(CatId),
    primary key(MedId,CatId)


    );


SELECT medicinecat.MedID AS 'MedId', medicine.Name AS 'Name', medicine.price AS 'Price', medicine.Quantity AS 'Quantity'
FROM medicinecat
JOIN medicine ON medicine.MedId = medicinecat.MedId
WHERE CatId in (1,2,3);

我得到的是所有至少有一个类别的药物，我知道这是因为 IN 子句， 我想要的是选择具有“所有类别”而不是其中一个类别的药物。

Answer 1

select
    m.MedID AS 'MedId',
    m.Name AS 'Name',
    m.price AS 'Price',
    m.Quantity AS 'Quantity'
from medicine as m
where
    exists (
        -- to get only those having all three categories
        select 1
        from medicinecat as mc
        where mc.CatId in (1,2,3) and mc.MedID = m.MedID 
        having count(*) = 3
    )

如果你能做一些通用的解决方案，比如

select
    m.MedID AS 'MedId',
    m.Name AS 'Name',
    m.price AS 'Price',
    m.Quantity AS 'Quantity'
from medicine as m
where
    not exists (
        select *
        from (values (1), (2), (3)) as tt(CatId)
            left outer join medicinecat as mc on mc.CatId = tt.CatId and mc.MedID = m.MedID
        where mc.CatId is null
    )

或

select
    m.MedID AS 'MedId',
    m.Name AS 'Name',
    m.price AS 'Price',
    m.Quantity AS 'Quantity'
from medicine as m
where
    exists (
        -- to get only those having all categories
        select 1
        from (values (1), (2), (3)) as tt(CatId)
            left outer join medicinecat as mc on mc.CatId = tt.CatId and mc.MedID = m.MedID
        having count(tt.CatId) = count(mc.CatId)
    )