选择以选择同时浏览过页面的用户

Question

我有一个表记录每个用户的页面浏览量：

+--------------+--------------+------+-----+---------+----------------+
| Field        | Type         | Null | Key | Default | Extra          |
+--------------+--------------+------+-----+---------+----------------+
| view_id      | int(11)      | NO   | PRI | NULL    | auto_increment | 
| page_id      | int(11)      | YES  | MUL | NULL    |                | 
| user_id      | int(11)      | YES  | MUL | NULL    |                | 
+--------------+--------------+------+-----+---------+----------------+

对于每对用户，我想生成一个他们都看过多少页面的计数。

我根本不知道该怎么做。 :)我使用的是mysql，以防它具有非标准功能，使它轻而易举。

Answer 1

select u1.user_id, u2.user_id, count(distinct u1.page_id) as NumPages
from logtable u1
  join
  logtable u2
  on u1.page_id = u2.page_id
  and u1.user_id < u2.user_id /* This avoids counting pairs twice */
group by u1.user_id, u2.user_id;

但是您应该考虑对此进行过滤...

（在上面编辑以放入u1.page_id，它原来只是page_id，对我来说真的很不好）

Answer 2

SELECT DISTINCT page_id
FROM logtable
WHERE user_id = 1 OR user_id = 2
GROUP BY page_id
HAVING COUNT(DISTINCT user_id) = 2

该表返回他们都已经查看过的所有页面。 如果要计数，则只需将此子查询并计数行即可。

SELECT COUNT(*) FROM (the query above) s;

更新，然后让所有用户对都可以进行更新。

SELECT u1.user_id, u2.user_id, COUNT(DISTINCT u1.page_id)
FROM logtable u1, logtable u2
WHERE u1.user_id < u2.user_id
  AND u1.page_id = u2.page_id
GROUP BY u1.user_id, u2.user_id

Answer 3

select a.user_id as user1, b.user_id as user2, count(distinct a.page_id) as views
from yourtable a, yourtable b
where a.page_id = b.page_id 
and a.user_id < b.user_id
group by a.user_id, b.user_id

将yourtable更改为表的名称..

Answer 4

对于users_id 100和200。

SELECT
  page_id
FROM table1
WHERE user_id IN (100, 200)
GROUP BY page_id
HAVING MAX(CASE WHEN user_id = 100 THEN 1 ELSE 0 END) = 1
  AND MAX(CASE WHEN user_id = 200 THEN 1 ELSE 0 END) = 1;