Question

表名：d_risposte id：index，唯一不相关回复：从1到4

id | reply | user
-----------------
  |   1   |  a34
  |   2   |  a34
  |   1   |  a34
  |   3   |  a34
  |   2   |  a34
  |   4   |  a34
  |   1   |  a34
  |   1   |  a34
  |   4   |  a55
  |   2   |  a55
  |   2   |  a55
  |   4   |  a71
  |   4   |  a71
  |   1   |  a71
  |   3   |  a71
  |   4   |  a71
  |   2   |  a71
  |   4   |  a71 
  |   4   |  a71 
  |   2   |  b22
  |   4   |  b22
  |   2   |  b22

我希望只保留以“a”开头且至少有7条回复的用户，然后我希望计算大多数重播1..4在这种情况下结果应该是：

   user | top_reply
   ----------------
   a34  |  1         (becouse the 1 reply appears 4 times)
   a71  |  4         (becouse the 4 reply appears 5 times)

（如果它是平局，那么重要）

所以我的查询是：

SELECT user FROM `d_risposte` WHERE user LIKE 'a%' group by user having count(*) > 7

表格结果应为：

user
----
a34
a71

然后我应该计算大部分回复：但我不知道如何做一个查询，我试着这样的......

SELECT reply, COUNT(reply) AS fooCount FROM d_risposte WHERE user ????????? GROUP BY reply ORDER BY COUNT(reply) DESC LIMIT 1

我无法弄明白:(

Answer 1

这是一个有趣的问题，试试这个：

SELECT DISTINCT t.user,
       (SELECT s.reply FROM d_risposte s
        WHERE s.user = t.user
        GROUP BY s.reply
        ORDER BY count(*) DESC limit 1) as top_reply
FROM d_risposte t
INNER JOIN(SELECT d.user FROM d_risposte d
           WHERE d.user like 'a%'
           GROUP BY d.user
           HAVING count(*) > 6) t2
 ON(t2.user = t.user)

这将基本为每个用户选择top_reply，然后加入表格以过滤那些记录少于7条并且不以a开头的人

Answer 2

您可以使用两个聚合执行此操作：

Select user,
      Substring_index(group_concat(reply order by cnt desc), ',', 1) as reply
From (select user, reply, count(*) as can't
      From d_riposte
      Group by user, reply) ur
Group by user
Having sum(cnt) >= 7

如何找到具有计数＆gt;的值的最多外观7

2 个答案: