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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, priceFROM   shop s1WHERE  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 (seeSection 13.2.10.7, “Correlated Subqueries”). Other possibilities for solving the problem are to use an uncorrelated subquery in theFROM clause or aLEFT JOIN.

Uncorrelated subquery:

SELECT s1.article, dealer, s1.priceFROM shop s1JOIN (  SELECT article, MAX(price) AS price  FROM shop  GROUP BY article) AS s2  ON s1.article = s2.article AND s1.price = s2.priceORDER BY article;

LEFT JOIN:

SELECT s1.article, s1.dealer, s1.priceFROM shop s1LEFT JOIN shop s2 ON s1.article = s2.article AND s1.price < s2.priceWHERE s2.article IS NULLORDER BY s1.article;

TheLEFT JOIN works on the basis that whens1.price is at its maximum value, there is nos2.price with a greater value and thus the correspondings2.article value isNULL. SeeSection 13.2.9.2, “JOIN Clause”.