SQL结果表，与第二个表的SET类型匹配

小编典典

SQL结果表，与第二个表的SET类型匹配

sql

下面的两个表不受任何类型的约束。

首先，我有一个名为以下表格subscription_plans：

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

我还有第二张桌子pricing_offers。的subscription_plan_ID是一种类型的SET，且只能包含匹配ID的所述的那值subscription_plans.ID（列从上面的表）。该表如下所示：

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

我试图做一个查询，以选择第一个表中的所有内容（所有字段*）和第二个表中的所有名称，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

这样的想法是，对于subscription_plans表中的每一行，它都应该看起来ID字段。然后走线槽第二个表，看看行包含在subscription_plan_ID中，ID该行的上方。将它们收集到字段调用器中p_o_name，并将其值插入匹配的响应行中。

我尝试这样做：

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：

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

注意：我得到所有行的响应，但是我只是将第一个放在此处以举例说明不同之处。

现在，尽管我可以在PHP页面上的响应中进行处理，但我想知道是否可以让数据库引擎输出所需的结果。我需要在表之间创建一种约束类型吗？如果是这样，我该怎么办？我将不胜感激能帮助我达到预期的输出结果（甚至是该问题的更好标题！）。

如果有任何不清楚的地方，请让我知道，我将予以澄清。

阅读 180

2021-03-17

共1个答案

小编典典

联结/相交表用法示例。

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，您可以从联结表中插入和删除（在此示例中，确实没有很好的更新过）。

干净和快速的连接，而不必弄乱没有索引的缓慢而笨拙的集

没有人可以再骑Wilbur the Poney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。

还有祝你好运！

2021-03-17