MySQL 8.4 Release Notes
5.6.4 包含特定列分组最大值的行
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;