按顺序获取数据库中的最后N行？

Question

假设我有以下数据库表:

 record_id | record_date | record_value
-----------+-------------+--------------
         1 | 2010-05-01  |       195.00
         2 | 2010-07-01  |       185.00
         3 | 2010-09-01  |       175.00
         4 | 2010-05-01  |       189.00
         5 | 2010-06-01  |       185.00
         6 | 2010-07-01  |       180.00
         7 | 2010-08-01  |       175.00
         8 | 2010-09-01  |       170.00
         9 | 2010-10-01  |       165.00

我想用record_date ASC排序的数据获取最后5行.这很容易做到:

SELECT * FROM mytable ORDER BY record_date ASC LIMIT 5 OFFSET 4

哪个会给我:

 record_id | record_date | record_value
-----------+-------------+--------------
         6 | 2010-07-01  |       180.00
         7 | 2010-08-01  |       175.00
         3 | 2010-09-01  |       175.00
         8 | 2010-09-01  |       170.00
         9 | 2010-10-01  |       165.00

但是当我不知道有多少记录并且无法计算4的幻数时,我该怎么做呢？

我已尝试过此查询,但如果记录少于5条,则会导致OFFSET为负,这是无效的:

SELECT * FROM mytable ORDER BY record_date ASC LIMIT 5 
    OFFSET (SELECT COUNT(*) FROM mytable) - 5;

那我该怎么做呢？

Answer 1

你为什么不以相反的方式订购？

SELECT * FROM mytable ORDER BY record_date DESC LIMIT 5;

如果您不想在应用程序中正确翻回,可以嵌套查询并翻转它们两次:

SELECT *
    FROM (SELECT * FROM mytable ORDER BY record_date DESC LIMIT 5)
    ORDER BY record_date ASC;

......结果证明这是一个相当便宜的操作.

Answer 2

这应该工作:

WITH t AS (
    SELECT * FROM mytable ORDER BY record_date DESC LIMIT 5
)
SELECT * FROM t ORDER BY record_date ASC;