MySQL Reference Manual for version 4.0.18.

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3.6.4 The Rows Holding the Group-wise Maximum of a Certain Field

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

In SQL-99 (and MySQL Version 4.1 or greater), the problem can be solved with a subquery like this:

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

In MySQL versions prior to 4.1, it's best do it in several steps:

  1. Get the list of (article,maxprice) pairs.
  2. For each article, get the corresponding rows that have the stored maximum price.

This can easily be done with a temporary table and a join:

 
CREATE TEMPORARY TABLE tmp (
        article INT(4) UNSIGNED ZEROFILL DEFAULT '0000' NOT NULL,
        price   DOUBLE(16,2)             DEFAULT '0.00' NOT NULL);

LOCK TABLES shop READ;

INSERT INTO tmp SELECT article, MAX(price) FROM shop GROUP BY article;

SELECT shop.article, dealer, shop.price FROM shop, tmp
WHERE shop.article=tmp.article AND shop.price=tmp.price;

UNLOCK TABLES;

DROP TABLE tmp;

If you don't use a TEMPORARY table, you must also lock the tmp table.

"Can it be done with a single query?"

Yes, but only by using a quite inefficient trick called the "MAX-CONCAT trick":

 
SELECT article,
       SUBSTRING( MAX( CONCAT(LPAD(price,6,'0'),dealer) ), 7) AS dealer,
  0.00+LEFT(      MAX( CONCAT(LPAD(price,6,'0'),dealer) ), 6) AS price
FROM   shop
GROUP BY article;

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

The last example can be made a bit more efficient by doing the splitting of the concatenated column in the client.



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