MySQL 9.0 Reference Manual  /  ...  /  The Rows Holding the Group-wise Maximum of a Certain Column

### 5.6.4 The Rows Holding the Group-wise Maximum of a Certain Column

Task: For each article, find the dealer or dealers with the most expensive price.

This problem can be solved with a subquery like this one:

``````SELECT article, dealer, price
FROM   shop s1
WHERE  price=(SELECT MAX(s2.price)
FROM shop s2
WHERE s1.article = s2.article)
ORDER BY article;

+---------+--------+-------+
| article | dealer | price |
+---------+--------+-------+
|    0001 | B      |  3.99 |
|    0002 | A      | 10.99 |
|    0003 | C      |  1.69 |
|    0004 | D      | 19.95 |
+---------+--------+-------+``````

The preceding example uses a correlated subquery, which can be inefficient (see Section 15.2.15.7, “Correlated Subqueries”). Other possibilities for solving the problem are to use an uncorrelated subquery in the `FROM` clause, a ```LEFT JOIN```, or a common table expression with a window function.

Uncorrelated subquery:

``````SELECT s1.article, dealer, s1.price
FROM shop s1
JOIN (
SELECT article, MAX(price) AS price
FROM shop
GROUP BY article) AS s2
ON s1.article = s2.article AND s1.price = s2.price
ORDER BY article;``````

`LEFT JOIN`:

``````SELECT s1.article, s1.dealer, s1.price
FROM shop s1
LEFT JOIN shop s2 ON s1.article = s2.article AND s1.price < s2.price
WHERE s2.article IS NULL
ORDER BY s1.article;``````

The `LEFT JOIN` works on the basis that when `s1.price` is at its maximum value, there is no `s2.price` with a greater value and thus the corresponding `s2.article` value is `NULL`. See Section 15.2.13.2, “JOIN Clause”.

Common table expression with window function:

``````WITH s1 AS (
SELECT article, dealer, price,
RANK() OVER (PARTITION BY article
ORDER BY price DESC
) AS `Rank`
FROM shop
)
SELECT article, dealer, price
FROM s1
WHERE `Rank` = 1
ORDER BY article;``````