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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
Say I have the id 32kJ. How would I get the next row or the previous one?
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1) You can use MAX or MIN along with OVER clause and add extra condition to it. The extra condition is "ROWS BETWEEN 1 PRECEDING AND 1 PRECEDING" which will fetch previous row value. Check this: SELECT *,MIN(JoiningDate) OVER (ORDER BY JoiningDate ROWS BETWEEN 1 PRECEDING AND 1 PRECEDING) AS EndDate.
One of the easiest ways, to compare this is using the lag function. The lag function will allow you to shift the rows downward so that you can view these rows as one observational row.
For example, by using the LEAD() function, from the current row, you can access data of the next row, or the row after the next row, and so on. The LEAD() function can be very useful for comparing the value of the current row with the value of the following row.
You need to use ORDER BY clause to get the second last row of a table in MySQL. The syntax is as follows. select *from yourTableName order by yourColumnName DESC LIMIT 1,1; To understand the above syntax, let us create a table.
This is what I use for finding previous/next records. Any column in your table can be used as the sort column, and no joins or nasty hacks are required:
Next record (date greater than current record):
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;
Previous record (date less than current record):
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;
Example:
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 |
+------+---------------+---------------------+
I realize that you are using MySQL, but just for reference, here is how you would do this using Oracle's analytic functions LEAD and LAG:
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
I guess there's a reason why Oracle costs money and MySQL is free... :-)
This page shows you how to emulate analytic functions in MySQL.
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