编写一个查询，可以显示哪些政党在所有选区都有候选人？

Question

所以我创建了 3 个 table 值

#创建派对table

DROP TABLE IF EXISTS parties,districts,candidates;
CREATE TABLE parties ( party char(12) NOT NULL, PRIMARY KEY (party) );
INSERT INTO parties VALUES ('Conservative'),('Liberal'), ('Socialist'),('Green'),('Libertarian');

#创建选区table

CREATE TABLE districts ( district char(10) DEFAULT NULL );
INSERT INTO districts VALUES ('Essex'), ('Malton'),('Riverdale'),('Guelph'),('Halton');

#创建候选人table

CREATE TABLE candidates ( id int(11) NOT NULL, name char(10) DEFAULT NULL, district char(10) DEFAULT NULL, party char(10) DEFAULT NULL, PRIMARY KEY (id) ); 
INSERT INTO candidates VALUES (1,'Anne Jones','Essex','Liberal'), (2,'Mary Smith','Malton','Liberal'), (3,'Sara Black','Riverdale','Liberal'), (4,'Paul Jones','Essex','Socialist'), 
(5,'Ed White','Essex','Conservative'), (6,'Jim Kelly','Malton','Liberal'), (7,'Fred Price','Riverdale','Socialist'), (8,'Bill Green','Guelph','Green'), 
(9,'Garth Adams','Halton','Libertarian'), (10,'Sam Adams','Guelph','Liberal'), (11,'Jill Mackay','Halton','Liberal');

现在想知道各区有哪些政党有候选人？

我已经这样做了，但我不确定这是否是正确的方法！！有人可以指导我吗？

select p.party 
from parties p 
    inner join candidates c on p.party = c.party
    inner join districts d on c.district = d.district;

Answer 1

您可以使用聚合。然后用having统计区数，看是否都包括在内：

select p.party
from parties p inner join
     candidates c
     on p.party = c.party
group by p.party
having count(distinct district) = (select count(*) from districts);

count(distinct)处理一个政党在一个选区内有多于一名候选人的情况。

Answer 2

双反加入（排除存在选区但没有该党候选人的政党：

SELECT * FROM parties p
WHERE NOT EXISTS (
        SELECT * FROM districts dx
        WHERE NOT EXISTS (
                SELECT * FROM candidates cx
                WHERE cx.district = dx.district
                AND cx.party = p.party
                )
        );

编写一个查询，可以显示哪些政党在所有选区都有候选人？

Write a query that can show which parties have candidates in all districts?

mysql

sql

relational-division