Question

我有两个表：

person

+-----+------------+---------------+
| id  | name       | address       |
+-----+------------+---------------+
| 1   | John Smith | 123 North St. |
| 2   | Joe Dirt   | 456 South St. |
+-----+------------+---------------+

person_fields

+-----+------------+-----------+-------+
| id  | type       | person_id | value |
+-----+------------+-----------+-------+
| 1   | isHappy    | 1         | 1     |
| 2   | hasFriends | 1         | 1     |
| 3   | hasFriends | 2         | 1     |

我想从person和isHappy为TRUE的hasFriends中选择所有人员。这是我尝试过的：

SELECT person.*
    FROM person
      INNER JOIN person_fields
        ON person.id = person_fields.person_id
    WHERE
      (person_fields.type = 'isHappy' AND person_fields.value IS TRUE)
      AND
      (person_fields.type = 'hasFriends' AND person_fields.value IS TRUE)

不幸的是，这不起作用，因为您无法在person_fields中拥有type = 'isHappy'和type = 'hasFriends'的单个记录。我无法OR这两个条件，因为那样会同时返回约翰·史密斯和乔·迪尔特，但我只想要约翰·史密斯，因为他是唯一一个快乐并同时有朋友的人。

有什么建议吗？预先感谢！

Answer 1

参加两次：

SELECT person.*
FROM person
INNER JOIN person_fields happy
        ON person.id = happy.person_id AND happy.type='isHappy' AND happy.value
INNER JOIN person_fields friends
        ON person.id = friends.person_id AND friends.type='hasFriends' AND friends.value

Answer 2

您可以两次加入person_fields，一次加入isHappy，一次加入hasFriends。

SELECT p.*
FROM person p 
INNER JOIN person_fields f1 ON p.id = f1.person_id
INNER JOIN person_fields f2 ON f1.person_id = f2.person_id 
WHERE f1.type = 'isHappy' AND f2.type = 'hasFriends'

我不确定value字段的位置，但是您可以抛出一个额外的条件，或者在需要时输入两个条件

AND f1.value = 1 AND f2.value = 1

Answer 3

标准解决方案如下：

SELECT person_id 
  FROM person_fields 
 WHERE type IN ('ishappy','hasfriends') 
 GROUP 
    BY person_id 
HAVING COUNT(1) = 2;

...其中'2'等于IN（）中的参数个数

请注意，这假设（person_id，type）是唯一的

Answer 4

如果您始终知道要检查的条件数，并且始终只想返回具有所有条件的person记录，则可以使用子查询并修改HAVING子句以选择type中person_fields的最大数量：

SELECT
  p.id,
  p.name,
  (SELECT COUNT(DISTINCT pf.type) FROM person_fields AS pf WHERE pf.person_id = p.id and pf.value = true) AS types
FROM
  person AS p
HAVING types = 2

结果：

id  name       types
1   John Smith 2

一对多关联SELECT记录两个或两个以上条件

4 个答案: