MySQL 8.0 Reference Manual Including MySQL NDB Cluster 8.0

*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 13.2.11.7, “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 13.2.10.2, “JOIN Syntax”.

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;