从MySQL中的最后一个条目中选择第二个

Question

问题 ：在指定时间段内，每个不同票证的倒数第二次需要

示例数据 ：

ticketTB
ticket createdate          status
111    2015-08-13 04:05:12 good
111    2015-08-13 04:04:12 bad
111    2015-08-13 04:03:12 good
115    2015-08-13 03:05:12 good
115    2015-08-13 03:04:12 bad
115    2015-08-13 03:03:12 good

查询 ：

SELECT ticket, status, createdate FROM ticketTB GROUP BY ticket ORDER BY createdate DESC LIMIT 1,1;

该查询提取：

ticket createdate          status
111    2015-08-13 04:04:12 bad

但是，添加许多其他票证只会使我需要它来评估每个不同票证的所有票务中的第二个倒数。

我希望查询返回：

ticket createdate          status
111    2015-08-13 04:04:12 bad
115    2015-08-13 03:04:12 bad

Answer 1

DROP TABLE IF EXISTS ticketTB;

CREATE TABLE ticketTB
(ticket INT NOT NULL
,createdate DATETIME NOT NULL
,status VARCHAR(12)
,PRIMARY KEY(ticket,createdate)
);

INSERT INTO ticketTB VALUES
(111    ,'2015-08-13 04:05:12','good'),
(111    ,'2015-08-13 04:04:12','bad'),
(111    ,'2015-08-13 04:03:12','good'),
(111    ,'2015-08-13 04:02:12','good'),
(115    ,'2015-08-13 03:05:12','good'),
(115    ,'2015-08-13 03:04:12','bad'),
(115    ,'2015-08-13 03:03:12','good'),
(115    ,'2015-08-13 03:02:12','good');

SELECT * FROM ticketTB;
+--------+---------------------+--------+
| ticket | createdate          | status |
+--------+---------------------+--------+
|    111 | 2015-08-13 04:02:12 | good   |
|    111 | 2015-08-13 04:03:12 | good   |
|    111 | 2015-08-13 04:04:12 | bad    |
|    111 | 2015-08-13 04:05:12 | good   |
|    115 | 2015-08-13 03:02:12 | good   |
|    115 | 2015-08-13 03:03:12 | good   |
|    115 | 2015-08-13 03:04:12 | bad    |
|    115 | 2015-08-13 03:05:12 | good   |
+--------+---------------------+--------+

SELECT x.* 
  FROM ticketTB x 
  JOIN ticketTB y 
    ON y.ticket = x.ticket 
   AND y.createdate >= x.createdate 
 GROUP 
    BY ticket
     , createdate 
HAVING COUNT(*) = 2;
+--------+---------------------+--------+
| ticket | createdate          | status |
+--------+---------------------+--------+
|    111 | 2015-08-13 04:04:12 | bad    |
|    115 | 2015-08-13 03:04:12 | bad    |
+--------+---------------------+--------+

类似于以下方法，通常会更快，尤其是在较大的数据集上。

SELECT ticket
     , createdate
     , status 
  FROM
     ( SELECT *
            , CASE WHEN @prev = ticket THEN @i:=@i+1 ELSE @i:=1 END rank
            , @prev := ticket prev 
         FROM ticketTB
            , (SELECT @prev:='',@i:=1) vars 
        ORDER 
           BY ticket
            , createdate DESC
     ) x 
 WHERE rank = 2;
+--------+---------------------+--------+
| ticket | createdate          | status |
+--------+---------------------+--------+
|    111 | 2015-08-13 04:04:12 | bad    |
|    115 | 2015-08-13 03:04:12 | bad    |
+--------+---------------------+--------+

Answer 2

尝试这个：

SELECT * FROM my_table ORDER BY rating DESC LIMIT 2,1