Question

这是我的表x

user | action

 1   |  A
 1   |  A
 1   |  B
 1   |  C
 1   |  c
 2   |  A
 2   |  B
 2   |  B

我想知道用户的前2个动作。意思是我希望结果为

1 | A, C
2 | A, B ( or B, A )

我尝试了很多方法，例如

 select user , action, COUNT(*) from table where user is NOT NULL group by user, action  order by user, COUNT(*);

但我无法得到理想的结果。

Answer 1

你可以这样做

SELECT user, SUBSTRING_INDEX(GROUP_CONCAT(action ORDER BY rank DESC, action), ',', 2) action
  FROM
(
  SELECT user, action, COUNT(*) rank
    FROM table1
   GROUP BY user, action
) q
 GROUP BY user

或

SELECT user, GROUP_CONCAT(action) action
  FROM
(  
  SELECT user, action, rank, @n := IF(@g = user, @n + 1, 1) rnum, @g := user
    FROM
  (
    SELECT user, action, COUNT(*) rank
      FROM table1
     GROUP BY user, action
  ) q CROSS JOIN (SELECT @n := 0, @g := NULL) i
   ORDER BY user, rank DESC, action
) q2
 WHERE rnum IN(1, 2)
 GROUP BY user

输出：

| USER | ACTION |
|------|--------|
|    1 |    A,C |
|    2 |    B,A |

这是 SQLFiddle 演示

Sql表示前2个可重复的值

1 个答案: