Question

我正在接收来自post的user_id 1和2。我希望像4这样的用户分享医院ID。

user_hopitals表

x = [[a b c d e f g e b]]
res, _ = string.gsub(x, "%a b", "z")
print(res)
-- z c d e f g z

医院表

 id | user_id | hospital_id
 1  |   1     |    4
 2  |   2     |    4
 3  |   1     |    5
 4  |   2     |    9

我想要像

这样的数据

id | name 
4  | abc hospital
5  | XYZ hospital
9  | def hospital

Answer 1

您可以使用SELF JOIN，例如：

SELECT h.id. h.name
FROM user_hospitals uh1 JOIN user_hospitals uh2 ON uh1.hospital_id = uh2.hospital_id
JOIN hospitals h ON uh1.hospital_id = h.id
WHERE uh1.user_id = 1 AND uh2.user_id = 2;

Answer 2

基本思想是聚合和having子句：

select uh.hospital_id, uh.name
from user_hospitals uh join
     hospitals h
     on uh.hopital_id = h.id
where uh.user_id in (1, 2)
group by uh.hospital_id, uh.name
having count(distinct uh.user_id) = 2;

MySQL与单表的相互条件查询

2 个答案: