mysql表绑定下一行和当前行

Question

我需要让下一行也绑定到我的查询响应。例如，

SELECT * FROM `ActivityLog` WHERE data like '%865814071038085%'

这会返回我需要的记录，使用它我还需要在结果集中显示下一行

我在这个查询中获得的表记录ID是，

1,
3,
5,
7

所以我需要一种方法来获取2,4,6,8的下一行并使查询输出

1,
2,
3,
4,
5,
6,
7,
8 
records.

如何使用mysql解释下一行？它总是当前行和下一个第一行。

CREATE TABLE IF NOT EXISTS `ActivityLog` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `activityAt` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
  `username` varchar(64) COLLATE utf8_unicode_ci DEFAULT NULL,
  `userId` int(11) DEFAULT NULL,
  `module` varchar(64) COLLATE utf8_unicode_ci DEFAULT NULL,
  `controller` varchar(64) COLLATE utf8_unicode_ci DEFAULT NULL,
  `action` varchar(64) COLLATE utf8_unicode_ci DEFAULT NULL,
  `data` text COLLATE utf8_unicode_ci NOT NULL,
  PRIMARY KEY (`id`),
  KEY `idxUserName` (`username`),
  KEY `idxUserId` (`userId`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci COMMENT='Application activity log' AUTO_INCREMENT=25390 ;

数据集，

INSERT INTO `ActivityLog` (`id`, `activityAt`, `username`, `userId`, `module`, `controller`, `action`, `data`) VALUES
(1, '2015-03-03 13:02:03', 'guest', NULL, '', 'Site', 'Login', 'Login access. IP:220.247.236.99'),
(2, '2015-03-03 13:02:08', 'QelasySuperAdmin', -1, '', 'Site', 'Login', 'Login success. IP: 220.247.236.99'),
(3, '2015-03-03 13:02:10', 'QelasySuperAdmin', -1, 'Log', 'Activity', 'Admin', 'Manage Activity Logs.'),
(4, '2015-03-03 13:02:17', 'QelasySuperAdmin', -1, 'Log', 'Activity', 'Admin', 'Manage Activity Logs.'),
(5, '2015-03-03 13:06:10', 'guest', NULL, 'Api', 'Default', 'DeviceUsage', ' Data:{"url":"http:\\/\\/www.googleadservices.com\\/pagead\\/aclk?sa=L&ai=CpOAR1LH1VPbZI4GOjAbY2ILQCeikiaQGwIKYmM8Bvtf9EwgAEAEgs_j-GCgDYIEDoAGwhoDQw&val=ChBhOTg3YjE3Y2E3MzM3NWYwEJC5jKcFGggTkt0OPC8P_iABKAAwwZ6E452z_YSAATiQuYynBUDMw9OnBQ&:","id":"1745"}'),
(6, '2015-03-03 13:06:10', 'guest', NULL, 'Api', 'Default', 'DeviceUsage', ' Response:{"code":200,"status":true,"data":[],"serverTime":"2015-03-03 13:06:10"}'),
(7, '2015-03-03 13:06:11', 'guest', NULL, 'Api', 'Default', 'DeviceUsage', ' Data:{"url":"about:blank","id":"1745"}'),
(8, '2015-03-03 13:06:11', 'guest', NULL, 'Api', 'Default', 'DeviceUsage', ' Response:{"code":200,"status":true,"data":[],"serverTime":"2015-03-03 13:06:11"}'),
(9, '2015-03-03 13:06:15', 'guest', NULL, 'Api', 'Default', 'DeviceUsage', ' Data:{"url":"http:\\/\\/clickserve","id":"1745"}'),
(10, '2015-03-03 13:06:15', 'guest', NULL, 'Api', 'Default', 'DeviceUsage', ' Response:{"code":200,"status":true,"data":[],"serverTime":"2015-03-03 13:06:15"}'))

更新

请检查此sql小提琴， http://sqlfiddle.com/#!9/b218a/1

到目前为止，没有一个答案对我有帮助。

EXPLAIN命令的输出。

Answer 1

试试这个。应该很接近。

select * from
  (
    SELECT * FROM `ActivityLog` WHERE data like '%865814071038085%'
    union
    SELECT * FROM `ActivityLog` where id in (SELECT id + 1 FROM `ActivityLog` WHERE data like '%865814071038085%') 
  ) q

演示：http://sqlfiddle.com/#!9/2b406/5

以防ids是非连续的，这是另一个：

select * from
  (
    SELECT * FROM `ActivityLog` WHERE data like '%FOO%'
    union
    SELECT * FROM `ActivityLog` where id in (SELECT min(a2.id)  FROM `ActivityLog` a1 inner join `activitylog` a2 on a1.id < a2.id WHERE a1.data like '%FOO%' group by a1.id) 
  ) q

更新小提琴：http://sqlfiddle.com/#!9/2b406/9

使用与另一个答案中的Anirban N相同的技术，但修改为仅在{NOT}下一个序列id时更新@pid的值，并且仅在搜索到的值时匹配，或者在下一个时匹配顺序ID。 （如果id序列中存在间隙，则不会获得下一行）

select if(@pid + 1 = id, id, @pid := id) id, 
       activityAt,
       username , 
       userId, 
       module,
       controller,
       action,
       data  
  from `ActivityLog` as a 
  where a.data like '%865814071038085%' 
    or id = @pid + 1;

final fiddle

Answer 2

可以将结果存储在MySql变量（@pid）中，并在下一个联合查询中使用它，

工作示例

编辑版本2 （SQL FIDDLE）

SET @pid:= 0;

SELECT @pid:=id, `activityAt` ,`username` , `userId` , `module` ,`controller`,`action` ,`data`  
FROM `ActivityLog` as a WHERE a.data like '%865814071038085%' OR @pid!=0;

Answer 3

这应该可以在下一个更高的Id上自行加入表格。

SELECT * FROM
  (
    SELECT * FROM `ActivityLog` 
    WHERE data LIKE '%865814071038085%'
    UNION
    SELECT * 
    FROM `ActivityLog` 
    WHERE id IN (
      SELECT t2.id 
      FROM `ActivityLog` t1
      JOIN `ActivityLog` t2
        ON t2.id = t1.id+1 
     ) 
)t
ORDER BY Id

<强> SQL FIDDLE