Question

我有一张表如下

id comm_id user_id ..

这将用户订阅存储到社区。现在我想查询这个表，以便只获取那些在2个用户之间相互作用的订阅（比如user_id：2和9）

给我sql查询 tablename：db_community_subscribers id，community_id，user_id

"20"      1         "2"
"28",  NULL         "2"
"31"   NULL,        "2"
"43",   4           "2"
"47     1             9
"57",   NULL,       "2"
"59",   "12",      "9"
"60     14          2
"62",  NULL         2

这些是用户2和9的订阅

我想只选择1和12，因为这些是常见的订阅

Answer 1

试试这个：

我认为现在可行了：

SELECT Comm_ID
FROM tableNAme
WHERE  user_id IN (2,9)
GROUP BY Comm_ID
HAVING COUNT(Comm_ID) > 1

或

SELECT DISTINCT d.Comm_ID FROM (SELECT Comm_ID FROM tableNAme WHERE user_id in (1,2)) d GROUP BY d.Comm_ID HAVING COUNT(d.Comm_ID) > 1

Answer 2

尝试以下

  SELECT group_concat(distinct comm_id)
  FROM table 
  WHERE user_id in(2, 9)
  group by user_id  having count(id) >1

OR：

  SELECT distinct comm_id
  FROM table 
  WHERE user_id in(2, 9)
  group by user_id  having count(id) >1

请做必要的更改如果您想要其他任何内容，请告诉我。

Answer 3

尝试：

select comm_id, group_concat(id) as subscription_ids
from subscriptions
where user_id in (2,9)
group by comm_id
having count(distinct user_id)=2

Answer 4

你试过一个简单的JOIN吗？

SELECT t1.`community_id`
FROM `tablename` t1
LEFT JOIN tablename t2 ON (t1.`community_id` = t2.`community_id`)
WHERE t1.`user_id` = 2
    AND t2.`user_id` = 9

对于每个user_id 2社区，应该找到匹配的用户9社区。

获得两个用户之间的相互订阅

4 个答案: