Question

我有一张桌子：

+------------+-----------------------------------------------------------------------------------+
| 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确定（最好只使用MySQL），如果自上一个type = LOGIN_FAIL以来或从表开始以来有{3}个连续记录。

例如

type = LOGIN_SUCCESS

会在

时返回

+----+---------------+---------+--------+---------------------+----+------------+
| 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      |
+----+---------------+---------+--------+---------------------+----+------------+

TRUE

会返回+----+---------------+---------+--------+---------------------+----+------------+ | 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 | +----+---------------+---------+--------+---------------------+----+------------+。是否可以通过简单的查询执行此操作，还是需要以某种脚本语言实现此检查？

编辑：我忘了提到这个查询必须限于某个user_id，但我认为这不会有问题。

否则，甚至更好，是否可以计算符合此条件的记录数（即自上次FALSE以来存在多少连续type = LOGIN_FAILED条记录）

Answer 1

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失败的数量。

Answer 2

希望这会对你有所帮助

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

MySQL检查3个或更多连续（特定）条目

2 个答案: