sql为当前行的下一行或上一行拉一行

Question

id    |  photo title     |  created_date

XEi43 |  my family       |  2009 08 04
dDls  |  friends group   |  2009 08 05
32kJ  |  beautiful place |  2009 08 06
EOIk  |  working late    |  2009 08 07 

说我有id 32kJ.我如何获得下一行或前一行？

Answer 1

这是我用于查找上一个/下一个记录的内容.表中的任何列都可以用作排序列,并且不需要连接或讨厌的黑客攻击:

下一条记录(日期大于当前记录):

SELECT id, title, MIN(created) AS created_date
FROM photo
WHERE created >
  (SELECT created FROM photo WHERE id = '32kJ')
GROUP BY created
ORDER BY created ASC
LIMIT 1;

以前的记录(日期少于当前记录):

SELECT id, title, MAX(created) AS created_date
FROM photo
WHERE created <
  (SELECT created FROM photo WHERE id = '32kJ')
GROUP BY created
ORDER BY created DESC
LIMIT 1;

例:

CREATE TABLE `photo` (
    `id` VARCHAR(5) NOT NULL,
    `title` VARCHAR(255) NOT NULL,
    `created` DATETIME NOT NULL,
    INDEX `created` (`created` ASC),
    PRIMARY KEY (`id`)
)
ENGINE = InnoDB;

INSERT INTO `photo` (`id`, `title`, `created`) VALUES ('XEi43', 'my family',       '2009-08-04');
INSERT INTO `photo` (`id`, `title`, `created`) VALUES ('dDls',  'friends group',   '2009-08-05');
INSERT INTO `photo` (`id`, `title`, `created`) VALUES ('32kJ',  'beautiful place', '2009-08-06');
INSERT INTO `photo` (`id`, `title`, `created`) VALUES ('EOIk',  'working late',    '2009-08-07');

SELECT * FROM photo ORDER BY created;
+-------+-----------------+---------------------+
| id    | title           | created             |
+-------+-----------------+---------------------+
| XEi43 | my family       | 2009-08-04 00:00:00 |
| dDls  | friends group   | 2009-08-05 00:00:00 |
| 32kJ  | beautiful place | 2009-08-06 00:00:00 |
| EOIk  | working late    | 2009-08-07 00:00:00 |
+-------+-----------------+---------------------+


SELECT id, title, MIN(created) AS next_date
FROM photo
WHERE created >
  (SELECT created FROM photo WHERE id = '32kJ')
GROUP BY created
ORDER BY created ASC
LIMIT 1;

+------+--------------+---------------------+
| id   | title        | next_date           |
+------+--------------+---------------------+
| EOIk | working late | 2009-08-07 00:00:00 |
+------+--------------+---------------------+

SELECT id, title, MAX(created) AS prev_date
FROM photo
WHERE created <
  (SELECT created FROM photo WHERE id = '32kJ')
GROUP BY created
ORDER BY created DESC
LIMIT 1;

+------+---------------+---------------------+
| id   | title         | prev_date           |
+------+---------------+---------------------+
| dDls | friends group | 2009-08-05 00:00:00 |
+------+---------------+---------------------+