从一个表中选择所有条目，在另一个表中有两个特定条目

Question

因此，我有2个这样定义的表：

CREATE TABLE tblPersons (
    id   INTEGER PRIMARY KEY AUTOINCREMENT,
    name TEXT
);

CREATE TABLE tblHobbies (
    person_id INTEGER REFERENCES tblPersons (id),
    hobby     TEXT
);

例如，我在tblPersons中添加了3个人：

1 | John
2 | Bob
3 | Eve

还有tblHobbies中的下一个爱好：

1 | skiing
1 | serfing
1 | hiking
1 | gunsmithing
1 | driving
2 | table tennis
2 | driving
2 | hiking
3 | reading
3 | scuba diving

我需要的是查询，该查询将向我返回具有一些特定爱好的人的列表。

我唯一能想到的是：

SELECT id, name FROM tblPersons
    INNER JOIN tblHobbies as hobby1 ON hobby1.hobby = 'driving'
    INNER JOIN tblHobbies as hobby2 ON hobby2.hobby = 'hiking'
    WHERE tblPersons.id = hobby1.person_id and tblPersons.id = hobby2.person_id;

但这很慢。 有没有更好的解决方案？

Answer 1

首先，您在tblHobbies上没有主键，这是查询缓慢（以及其他问题）的原因之一。 另外，您应该考虑在tblHobbies.hobby上创建索引。

其次，我建议您创建第三个表以证明模型中存在N：N基数，并避免多余的爱好。 就像是：

--Person
CREATE TABLE tblPersons (
    id   INTEGER PRIMARY KEY AUTOINCREMENT,
    name TEXT
);

--Hobby
CREATE TABLE tblHobbies (
    id INTEGER PRIMARY KEY AUTOINCREMENT,
    hobby TEXT
);

--Associative table between Person and Hobby
CREATE TABLE tblPersonsHobbies (
    person_id INTEGER REFERENCES tblPersons (id),
    hobby_id INTEGER REFERENCES tblHobbies (id),
    PRIMARY KEY (person_id, hobby_id)
);

添加一个额外的表，但这是值得的。

--Query on your current model
SELECT id, name FROM tblPersons
    INNER JOIN tblHobbies as hobby1 ON  tblPersons.id = hobby1.person_id
    WHERE hobby1.hobby IN ('driving', 'hiking');

--Query on suggested model
SELECT id, name FROM tblPersons
    INNER JOIN tblPersonsHobbies as personsHobby ON  tblPersons.id = personsHobby.person_id
    INNER JOIN tblHobbies as hobby1 ON hobby1.id = personsHobby.hobby_id
        WHERE hobby1.hobby IN ('driving', 'hiking');

Answer 2

您可以汇总兴趣爱好表以同时获得两个兴趣爱好的人：

select person_id
from tblhobbies
group by person_id
having count(case when hobby = 'driving' then 1 end) > 0
   and count(case when hobby = 'hiking' then 1 end) > 0

最好使用WHERE子句来限制记录的读取：

select person_id
from tblhobbies
where hobby in ('driving', 'hiking')
group by person_id
having count(distinct hobby) =2

（不过，表中的person + hobby应该有一个唯一的约束。然后，您可以删除DISTINCT 。正如我在评论部分中所说的，它甚至应该是person_id + hobby_id，并带有一个单独的hobby表。编辑：糟糕，我本应该阅读其他答案的。Michal已经在三个小时前提出了这个数据模型：-)

如果需要名称，请从人员表中选择，在上面的查询中找到ID：

select id, name
from tblpersons
where id in
(
  select person_id
  from tblhobbies
  where hobby in ('driving', 'hiking')
  group by person_id
  having count(distinct hobby) =2
);

使用更好的数据模型，您将取代

  from tblhobbies
  where hobby in ('driving', 'hiking')
  group by person_id
  having count(distinct hobby) =2

与

  from tblpersonhobbies
  where hobby_id in (select id from tblhobbies where hobby in ('driving', 'hiking'))
  group by person_id
  having count(*) =2