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MySQL 5.7 Reference Manual  /  ...  /  The Rows Holding the Group-wise Maximum of a Certain Column

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.

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, “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
  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;


SELECT s1.article,, 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, “JOIN Syntax”.

User Comments
User comments in this section are, as the name implies, provided by MySQL users. The MySQL documentation team is not responsible for, nor do they endorse, any of the information provided here.
  Posted by Francisco Tirado on December 30, 2010
O T: Although that would be awesome, it did not work for me and I couldn't find it in the manual.
  Posted by Deon Kuhn on February 9, 2012
In my particular case I was only working with one record, and the correlated sub query solution proved 50% faster than the 'group by' trick. I am getting the most recent record to be altered by user 6 by way of a main table and changes log.

FROM main_table iA
JOIN log_table iC ON iA.tid = iC.tid AND iC.userid = 6
WHERE iC.status = 1 AND = ( SELECT MAX( id ) `scmax` FROM log_table WHERE iA.transactionID = trxid )

was faster than

FROM main_table iA
JOIN ( SELECT id, tid, userID FROM ( SELECT id, tid, userID FROM log_table WHERE userID = 6 ORDER BY id DESC ) iiA GROUP BY tid ) iC ON iA.tid = iC.tid AND iC.status = 1

  Posted by Daniel Bermudez on November 7, 2012
I am using this as a tutorial and tried doing it by myself before seeing the "answer" and I came up with this:

WHERE price IN (SELECT MAX(price)
FROM shop GROUP BY article);

I think is very straightforward and easy to understand but I might be missing something.

Why nobody recommended this?
  Posted by Thibault Delor on November 15, 2012
@Daniel Bermudez :
You can find your query there :

"Why nobody recommended this?" :
1) it doesn't do what we want there
2) it is slow
  Posted by Predrag Bradaric on January 25, 2013
I agree with Robin Palotai. Example given by Kasey Speakman is "unreliable" in the long run.
As stated in MySQL reference manual ( values chosen by GROUP BY extension are indeterminate - it is not "guaranteed" that first row (top->down) with distinct value will be chosen in the final result.
Granted, tests have shown that (at the moment) MySQL server will choose first row (top->down) that it encounters but that can break in any future update.
  Posted by John Pratt on March 5, 2014
Perhaps a future version of MySQL will no longer be "indeterminate". It would be easier to write this query if the Select would reliably always select the top row.
  Posted by Alex Wilding on December 29, 2015
The example given by Kasey Speakman appears to fail when you change it to try to find the minimum prices by replacing "desc" with "asc". i.e;

FROM shop
ORDER BY price ASC) AS s
GROUP BY article

does not produce the output you may expect.
  Posted by Paul Macey on January 4, 2017
The solution suggestion by Kasey Speakman worked for me with a table of ~20 million rows and 7 columns - took 2+ minutes to run. However, the JOIN & LEFT JOIN suggestions ran for hours without completing.
  Posted by Andras Toth on January 7, 2017
The latest release doesn't seem to support the "GROUP BY" trick anymore, which is a performance issue (see earlier comments on slower JOIN queries) but also means the need of far more complex queries. Consider this:

FROM table
GROUP BY t.item;

This query results a table in which you will find a single row for every item with the latest date, but for those items which have multiple rows with the latest date, it provides you the one with the highest id value.

With "GROUP BY" trick adding a secondary group-wise maximum was so easy, you had to simply extend your ORDER BY clause with another ranking option.

With the JOIN solutions mentioned in the manual, the only way to do this, is to create an inner query for the primary maximum then using an outer query for the secondary maximum.

SELECT t12.item,t12.price,,
(SELECT t1.item,t1.price,,
FROM table t1
LEFT JOIN table t2 ON t1.item=t2.item and<
WHERE is NULL) t12
LEFT JOIN table t3 ON t12.item = t3.item AND AND<

Now compare it with the GROUP BY trick. MAX() based solutions are probably even more complicated. MySQL should really work out a more sophisticated solution for this. (I could mention other SQL DBMSs.)
  Posted by Richard JI on January 21, 2017
The query below is similar to "Uncorrelated subquery" but is more straightforward -

select * from shop
where (article, price) in
(select article, max(price) from shop group by article);
  Posted by Rick James on May 24, 2018
For scaling... Beware of some techniques; they are Order(N) or even Order(N*N), where N is the number of input rows. The Order(N*N) are usually recognizable by "... LEFT JOIN... < ..."

This link gives two techniques that scale better:

It also provide efficient code for "Top-N in each Group".
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