Tabela wyników SQL, dopasowanie w drugiej tabeli typu SET

Przykład użycia tabeli skrzyżowań/skrzyżowań.

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

Zasadniczo wstawiasz i usuwasz z tabeli skrzyżowań (w tym przykładzie nie ma sensu aktualizować).

Czyste i szybkie łączenia bez bałaganu z wolnymi, nieporęcznymi zestawami bez indeksów

Nikt już nie może jeździć na kucyku Wilbur? Następnie

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

Zapytaj, jeśli masz jakieś pytania.

I powodzenia!