sql为当前行的下一行或上一行提取一行

小编典典

sql为当前行的下一行或上一行提取一行

mysql

id | 照片标题| 创建日期

XEi43 | 我的家人 2009 08 04
dDls | 朋友组| 2009 08 05
32kJ | 美丽的地方 2009 08 06
EOIk | 工作到很晚 2009 08 07

说我有身份证32kJ。我将如何获得下一行或上一行？

阅读 1353

2020-05-17

共1个答案

小编典典

这就是我用来查找上一个/下一个记录的方法。表格中的任何列都可以用作排序列，并且不需要联接或讨厌的技巧：

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;

上一个记录（日期小于当前记录）：

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

2020-05-17