id | 照片标题 | 创建日期
XEi43 | 我的家庭 | 2009年08月04日 dDls | 朋友团体 | 2009年08月05日 32kJ | 美丽的地方 | 2009年08月06日 EOIk | 工作到深夜 | 2009年08月07日
假设我有一个id为32kJ
,如何获取上一行或下一行?
id | 照片标题 | 创建日期
XEi43 | 我的家庭 | 2009年08月04日 dDls | 朋友团体 | 2009年08月05日 32kJ | 美丽的地方 | 2009年08月06日 EOIk | 工作到深夜 | 2009年08月07日
假设我有一个id为32kJ
,如何获取上一行或下一行?
这是我用来查找前/后记录的方法。 表中的任意列都可以用作排序列,而不需要连接或使用复杂的技巧:
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;
之前的记录(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;
示例:
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 |
+------+---------------+---------------------+
select empno, ename, job,
lag(ename, 1) over (order by ename) as the_guy_above_me,
lead(ename, 2) over (order by ename) as the_guy_two_rows_below_me
from emp
order by ename
您是否想按日期获取下一行/上一行数据?如果是这样,您可以这样做:
select MyTable.*
from MyTable
join
(select id
from MyTable
where created_date < (select created_date from MyTable where id = '32kJ')
order by created_date desc, id desc
limit 1
) LimitedTable on LimitedTable.id = MyTable.fund_id;
使用麦克的MAX/MIN技巧,我们可以为各种事情制作上一个/下一个跳转。这个msAccess示例将返回股票市场数据表中每个记录的前一次收盘价。注意:'<='是为周末和节假日准备的。
SELECT
tableName.Date,
tableName.Close,
(SELECT Close
FROM tableName
WHERE Date = (SELECT MAX(Date) FROM tableName
WHERE Date <= iJoined.yesterday)
) AS previousClose
FROM
(SELECT Date, DateAdd("d",-1, Date) AS yesterday FROM tableName)
AS iJoined
INNER JOIN
tableName ON tableName.Date=iJoined.Date;
...'yesterday'演示了如何使用函数(Date-1)进行跳转;我们也可以简单地使用...
(SELECT Date FROM tableName) AS iJoined
/* previous record */
(SELECT MAX(Date) FROM tableName WHERE Date < iJoined.Date)
/* next record */
(SELECT MIN(Date) FROM tableName WHERE Date > iJoined.Date)
诀窍是我们可以使用 MAX\MIN 和 jump function() 来处理前面和后面 # 个某物
with yourresult as
(
select id, photo_title, created_date, ROW_NUMBER() over(order by created_date) as 'RowNum' from your_table
)
-- Previous
select * from yourresult where RowNum = ((select RowNum from yourresult where id = '32kJ') -1)
-- Next
select * from yourresult where RowNum = ((select RowNum from yourresult where id = '32kJ') +1)
有用吗?
CREATE SCHEMA temp
create table temp.emp (id integer,name varchar(50), salary varchar(50));
insert into temp.emp values(1,'a','25000');
insert into temp.emp values(2,'b','30000');
insert into temp.emp values(3,'c','35000');
insert into temp.emp values(4,'d','40000');
insert into temp.emp values(5,'e','45000');
insert into temp.emp values(6,'f','20000');
select * from temp.emp
SELECT
current.id, current.name, current.salary,
case
when current.id = 1 then current.salary
else
case
when current.salary > previous.salary then previous.salary
else current.salary
end
end
FROM
temp.emp AS current
LEFT OUTER JOIN temp.emp AS previous
ON current.id = previous.id + 1