6

我有一个下表：

+------------+-----------------------------------------------------------------------------------+
| Field      | Type                                                                              |
+------------+-----------------------------------------------------------------------------------+
| id         | int(10) unsigned                                                                  |
| type       | enum('REGISTER','ACTIVATE','LOGIN_SUCCESS','LOGIN_FAIL','LOGOUT','LOCK','UNLOCK') |
| user_id    | int(10) unsigned                                                                  |
| mod_id     | int(10) unsigned                                                                  |
| date       | timestamp                                                                         |
| ip         | int(10) unsigned                                                                  |
| user_agent | text                                                                              |
+------------+-----------------------------------------------------------------------------------+

我试图以最简单的方式（最好只使用 MySQL）确定type = LOGIN_FAIL自上次type = LOGIN_SUCCESS或自表开始以来是否有 3 个或更多连续记录。

例如

+----+---------------+---------+--------+---------------------+----+------------+
| id | type          | user_id | mod_id | date                | ip | user_agent |
+----+---------------+---------+--------+---------------------+----+------------+
|  6 | LOGIN_SUCCESS |       3 |   NULL | 2012-07-21 14:08:32 |  0 | agent      |
|  7 | LOGIN_FAIL    |       3 |   NULL | 2012-07-21 14:08:32 |  0 | agent      |
|  8 | LOGIN_FAIL    |       3 |   NULL | 2012-07-21 14:08:32 |  0 | agent      |
|  9 | LOGIN_FAIL    |       3 |   NULL | 2012-07-21 14:08:32 |  0 | agent      |
+----+---------------+---------+--------+---------------------+----+------------+

会回来TRUE，而

+----+---------------+---------+--------+---------------------+----+------------+
| id | type          | user_id | mod_id | date                | ip | user_agent |
+----+---------------+---------+--------+---------------------+----+------------+
|  6 | LOGIN_FAIL    |       3 |   NULL | 2012-07-21 14:08:32 |  0 | agent      |
|  7 | LOGIN_FAIL    |       3 |   NULL | 2012-07-21 14:08:32 |  0 | agent      |
|  8 | LOGIN_SUCCESS |       3 |   NULL | 2012-07-21 14:08:32 |  0 | agent      |
|  9 | LOGIN_FAIL    |       3 |   NULL | 2012-07-21 14:08:32 |  0 | agent      |
+----+---------------+---------+--------+---------------------+----+------------+

会回来FALSE的。是否可以通过简单的查询来做到这一点，还是我需要用某种脚本语言来实现这个检查？

编辑：我忘了提到这个查询必须限制为某个 user_id 但我认为这不会是一个问题。

否则，或者甚至更好，是否可以计算有多少记录符合此标准（即type = LOGIN_FAILED自 last 以来存在多少连续记录type=LOGIN_SUCCESS）

4

2 回答 2

2 
SELECT COUNT(*) FROM `table` 
WHERE 
    id > 
        (IFNULL(
            (SELECT id 
            FROM `table` 
            WHERE `type`='LOGIN_SUCCESS' 
            ORDER BY id DESC 
            LIMIT 1),0
        )
    AND `type`='LOGIN_FAIL'

将获得自上次成功以来的失败次数。

于 2012-07-21T13:35:14.473 回答
2

希望对你有帮助

SELECT IF(COUNT(a.id)>=3, TRUE, FALSE) AS fresult FROM last_login AS a, 
(
        SELECT COUNT( b.id ) AS cnt, MAX( b.id ) AS maxid FROM last_login AS b 
        WHERE b.login_type =  'LOGIN_SUCCESS' 
) AS c  

WHERE a.id>c.maxid OR c.cnt=0
GROUP BY a.login_type
于 2012-07-21T14:06:25.480 回答