Question

您好我需要获取与列出的所有groupid匹配的行

SELECT user_id,group_id 
FROM group_privilege_details g
WHERE g.group_id in (102,101)

如果任何一个groupid匹配，这将返回我。但是，我需要在列表中提及所有groupid的userid。

Answer 1

假设您不能拥有重复的user_id / group_id组合：

SELECT user_id,count(group_id)
 FROM group_privilege_details g
WHERE g.group_id in (102,101) 
GROUP BY user_id
HAVING count(group_id) = 2

Answer 2

以下是Steven对通用数组的查询的变体：

SELECT user_id
FROM   group_privilege_details
WHERE  group_id = ANY(my_array)
GROUP  BY 1
HAVING count(*) = array_length(my_array, 1)

只要满足这些要求（问题中没有提到），就可以工作：

（user_id，group_id）在group_privilege_details中是唯一的。
数组只有1维
基本数组元素是唯一的

通用解决方案，无论这些前提条件如何都可以使用

WITH   ids AS (SELECT DISTINCT unnest(my_array) group_id)
SELECT g.user_id
FROM   (SELECT user_id, group_id FROM group_privilege_details GROUP BY 1,2) g
JOIN   ids USING (group_id)
GROUP  BY 1
HAVING count(*) = (SELECT count(*) FROM ids)

unnest()为每个基本元素生成一行。 DISTINCT删除了可能的欺骗行为。子选择对表格也是如此。

此类查询的广泛选项列表：How to filter SQL results in a has-many-through relation

Answer 3

请找到我的已解决的查询：

select user_id,login_name from user_info where user_id in (
SELECT user_id  FROM 
group_privilege_details g WHERE g.group_id in 
(select group_id from group_privilege_details g,user_info u where u.user_id=g.user_id
and login_name='123')  
GROUP BY user_id HAVING count(group_id) = (select count(group_id) 
from group_privilege_details g,user_info u where u.user_id=g.user_id
and login_name='123') ) and login_name!='123'

检索与列出的所有值匹配的行

3 个答案: