Question

任何类型的约束都不喜欢以下两个表。

首先，我有一个名为subscription_plans的表，如下所示：

name   | price | ID
-------------------
plan_A | 9.99  | 1
Plan_B | 19.99 | 2
plan_C | 29.99 | 3

我有一个名为pricing_offers的第二个表。 subscription_plan_ID属于SET类型，只能包含与subscription_plans.ID的ID相匹配的值（上表中的列）。此表如下所示：

p_o_name      | subscription_plan_ID | ID
-----------------------------------------
free donuts   | 1                    | 1
extra sauce   | 1,2,3                | 2
pony ride     | 3                    | 3
bus fare -50% | 1,2,3                | 4

我正在尝试执行查询以从第一个表中选择所有内容（所有字段*），并且第二个表中的所有名称和结果行应如下所示：

name   | price | p_o_name                                | ID
-------------------------------------------------------------
plan_A | 9.99  | free donuts, extra sauce, bus fare -50% | 1
Plan_B | 19.99 | extra_sauce, bus fare -50%              | 2
plan_C | 29.99 | extra_sauce, pony ride, bus fare -50%   | 3

对于subscription_plans表中的每一行，它应该看起来ID字段。然后转到第二个表格，看看上面一行的subscription_plan_ID，ID中包含哪些行。将它们收集到字段调用方p_o_name中，并将其值插入匹配的响应行。

我试过这样做：

SELECT subscription_plans.*, pricing_offers.name
FROM subscription_plans INNER JOIN pricing_offers ON
FIND_IN_SET(subscription_plans.ID,subscription_plan_ID)

但我得到的不是：

plan_A | 9.99  | free donuts, extra sauce, bus fare -50% | 1

这样：

plan_A | 9.99  | free donuts   | 1
plan_A | 9.99  | extra sauce   | 1
plan_A | 9.99  | bus fare -50% | 1

注意：我收到所有行的回复，但我只是把第一个放在这里来举例说明差异。

现在，虽然我可以在我的PHP页面上的响应中进行处理，但我很想知道我是否让数据库引擎输出我想要的结果。我是否需要在表之间创建一种约束？如果是这样我怎么办？如果有任何帮助可以帮助我获得我提供的输出结果（甚至是问题的更好标题，我将不胜感激！）。

如果有任何不明确的地方，请告诉我，我会澄清它们。

Answer 1

交叉点/交叉表用法示例。

create table subscription_plans
(
    id int not null auto_increment primary key, -- common practice
    name varchar(40) not null,
    description varchar(255) not null,
    price decimal(12,2) not null
    -- additional indexes:
);

create table pricing_offers
(
    id int not null auto_increment primary key, -- common practice
    name varchar(40) not null,
    description varchar(255) not null
    -- additional indexes:
);

create table so_junction
(   -- intersects mapping subscription_plans and pricing_offers
    id int not null auto_increment primary key, -- common practice
    subId int not null,
    offerId int not null,

    -- row cannot be inserted/updated if subId does not exist in parent table
    -- the fk name is completely made up
    -- parent row cannot be deleted and thus orphaning children
    CONSTRAINT fk_soj_subplans 
        FOREIGN KEY (subId)
        REFERENCES subscription_plans(id),

    -- row cannot be inserted/updated if offerId does not exist in parent table
    -- the fk name is completely made up
    -- parent row cannot be deleted and thus orphaning children
    CONSTRAINT fk_soj_priceoffer 
        FOREIGN KEY (offerId)
        REFERENCES pricing_offers(id),

    -- the below allows for only ONE combo of subId,offerId
    CONSTRAINT soj_unique_ids unique (subId,offerId)
    -- additional indexes:
);

insert into subscription_plans (name,description,price) values ('plan_A','description',9.99);
insert into subscription_plans (name,description,price) values ('plan_B','description',19.99);
insert into subscription_plans (name,description,price) values ('plan_C','description',29.99);
select * from subscription_plans;

insert into pricing_offers (name,description) values ('free donuts','you get free donuts, limit 3');
insert into pricing_offers (name,description) values ('extra sauce','extra sauce');
insert into pricing_offers (name,description) values ('poney ride','Free ride on Wilbur');
insert into pricing_offers (name,description) values ('bus fare -50%','domestic less 50');

select * from pricing_offers;

insert so_junction(subId,offerId) values (1,1); -- free donuts to plans
insert so_junction(subId,offerId) values (1,2),(2,2),(3,2); -- extra sauce to plans
insert so_junction(subId,offerId) values (3,3); -- wilbur
insert so_junction(subId,offerId) values (1,4),(2,4),(3,4); -- bus to plans
select * from so_junction;

-- try to add another of like above to so_junction
-- Error Code 1062: Duplicate entry

-- show joins of all
select s.*,p.*
from subscription_plans s
join so_junction so
on so.subId=s.id
join pricing_offers p
on p.id=so.offerId
order by s.name,p.name

-- show extra sauce intersects
select s.*,p.*
from subscription_plans s
join so_junction so
on so.subId=s.id
join pricing_offers p
on p.id=so.offerId
where p.name='extra sauce'
order by s.name,p.name

基本上你是从联结表中插入和删除的（在这个例子中没有真正的更新）。

清洁和快速连接，而不必混淆没有索引的缓慢，笨拙的集合

没有人可以再乘坐威尔伯的Poney了吗？然后

delete from so_junction
where offerId in (select id from pricing_offers where name='poney ride')

询问您是否有任何问题。

祝你好运！

SQL结果表，匹配第二个表的SET类型

1 个答案: