Question

Using MYSQL I would like to refactor the following SELECT statement to return the entire record containing the newest invoice_date:

> SELECT id, invoice, invoice_date
  FROM invoice_items
  WHERE lot = 1047

id    invoice_id   invoice_date
-----------------------------------
3235    1047         2009-12-15 11:40:00
3295    1047         2009-12-15 16:00:00
3311    1047         2009-12-15 09:30:00
3340    1047         2009-12-15 13:50:00

Using the MAX() aggregate function and the GROUP BY clause gets me part of the way there:

> SELECT id, invoice_id, max(invoice_date)
  FROM invoice_items
  WHERE invoice_id = 1047
  GROUP BY invoice_id


id    invoice_id   invoice_date
-----------------------------------
3235    1047         2009-12-15 16:00:00

Notice that the query appears to get the MAX(invoice_date) correctly, but the id returned (3235) is not the id of the record containing the MAX(invoice_date) (3295) it is the id of the first record in the initial query.

How do I refactor this query to give me the the entire record that contains the MAX(invoice_date)?

The solution must use the GROUP BY clause, because I need to get newest invoice_date for each invoice.

Answer 1

This is the often-repeated “greatest-n-per-group” problem.

Here’s how I would solve it in MySQL:

SELECT i1.*
FROM invoice_items i1
LEFT OUTER JOIN invoice_items i2
  ON (i1.invoice_id = i2.invoice_id AND i1.invoice_date < i2.invoice_date)
WHERE i2.invoice_id IS NULL;

Explanation: for each row i1, try to find a row i2 with the same invoice_id and a greater date. If none are found (i.e. i2 is all nulls because of the outer join), then i1 must be the row with the greatest date for its invoice_id.

This solution using join tends to work better for MySQL, which is weak when optimizing both GROUP BY and subqueries.