Get the last N rows in the database in order?

Posted by Kristopher on Stack Overflow See other posts from Stack Overflow or by Kristopher
Published on 2010-04-08T02:04:02Z Indexed on 2010/04/08 2:13 UTC
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Let's say I have the following database table:

 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

I want to grab the last 5 rows with the data ordered by record_date ASC. This is easy to do with:

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

Which would give me:

 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

But how do I do this when I don't know how many records there are and can't compute the magic number of 4?

I've tried this query, but if there are less than 5 records, it results in a negative OFFSET, which is invalid:

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

So how do I accomplish this?

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