Question

所以，我有两个像这样定义的表：

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

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

例如，我有3个人加入 tblPersons ：

1 | John
2 | Bob
3 | Eve

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，它会返回一份有几个特定爱好的人的名单。

我唯一想出的就是：

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;

但它很慢。没有更好的解决方案吗？

Answer 1

首先，您在tblHobbies上没有主键，这是导致查询速度缓慢（以及其他问题）的一个原因。您还应考虑在tblHobbies.hobby上创建索引。

其次，我建议您创建第三个表来证明模型中存在N：N基数并避免多余的爱好。类似的东西：

--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，但它是值得的。

--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

您可以聚合爱好表以获得具有两种爱好的人：

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

或者更好地使用WHERE子句限制记录：

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

（表中应该有对人+爱好的唯一约束。然后你可以删除DISTINCT。正如我在评论部分中所说，它甚至应该是person_id + hobby_id，并且有一个单独的爱好编辑：哎呀，我应该读另一个答案.Michal已经在三小时前提出了这个数据模型： - ）

如果您需要名称，请从人员表中选择您在上述查询中找到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，您可以替换

  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

从一个表中选择所有条目，该表在另一个表中具有两个特定条目

2 个答案: