Question

这是查询

SELECT `mpv`.`member_id` as member_id, `screen_name`, `viewed_url`, 
    count(viewed_url) as viewed_url_cnt
FROM (`tracker` mpv)
JOIN `members` m ON `mpv`.`member_id`=`m`.`member_id`
GROUP BY `viewed_url`, `member_id`

它返回

+-----------+-------------+------------+----------------+
| member_id | screen_name | viewed_url | viewed_url_cnt |
+-----------+-------------+------------+----------------+
|         1 | admin       | /          |              1 |
|         1 | admin       | /1         |              1 |
|         1 | admin       | /2         |              1 |
|         1 | admin       | 2          |              1 |
|         1 | admin       | 3          |              2 |
|         2 | test        | 4          |              1 |
+-----------+-------------+------------+----------------+
6 rows in set (0.12 sec)

我需要获取viewed_url_cnt分组viewed_url member_id +-----------+-------------+------------+----------------+ | member_id | screen_name | viewed_url | viewed_url_cnt | +-----------+-------------+------------+----------------+ | 1 | admin | 3 | 2 | | 2 | test | 4 | 1 | +-----------+-------------+------------+----------------+ 6 rows in set (0.12 sec)的所有记录。所以基于顶部的结果，我想得到

{{1}}

由于

Answer 1

我可能会误解你想要的内容，但解决方案是使用内部查询为每个viewed_url_cnt确定适当的member_id，然后JOIN到您的表：

SELECT *
FROM my_qry outer
JOIN
(SELECT member_id, MAX(viewed_url_cnt) AS max_cnt
FROM my_qry
GROUP BY member_id) inner 
ON outer.member_id = inner.member_id AND outer.viewed_url_cnt = inner.max_cnt

（此处my_qry是您用于生成原始表的查询。）

Answer 2

我想以下是可能的：

SELECT * FROM (SELECT * FROM table GROUP BY name) as foo GROUP BY dept

用查询替换内部查询，然后您可以通过member_id获取max（seen_url_cnt）组。
如果失败，请告诉我..

Answer 3

试试这个：

select s1.member_id, s1.screen_name, s1.viewed_url, s1.viewed_url_cnt from (
  select t1.member_id, t1.screen_name, t1.viewed_url, count(*) viewed_url_cnt
  from tracker t1
  group by t1.member_id, t1.screen_name, t1.viewed_url
) as s1
join (
  select s2.member_id, max(viewed_url_cnt) as viewed_url_max from (
    select t1.member_id, count(*) viewed_url_cnt
    from tracker t1
    group by t1.member_id, t1.viewed_url
  ) as s2
  group by s2.member_id
) as s3
on s1.member_id = s3.member_id and s1.viewed_url_cnt = s3.viewed_url_max

按最大计数（字段）过滤结果

3 个答案: