首先把官网示例拿出来：
连接查询比子查询性能更好
3.6.4 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.
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 13.2.10.7, “Correlated Subqueries”). Other possibilities for solving the problem are to use an uncorrelated subquery in the FROM clause or a LEFT JOIN.
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 13.2.9.2, “JOIN Clause”.