Question

Consider this table (from http://www.tizag.com/mysqlTutorial/mysqlmax.php):

Id     name               type     price 
123451 Park's Great Hits  Music    19.99 
123452 Silly Puddy        Toy      3.99 
123453 Playstation        Toy      89.95 
123454 Men's T-Shirt      Clothing 32.50 
123455 Blouse             Clothing 34.97 
123456 Electronica 2002   Music    3.99 
123457 Country Tunes      Music    21.55 
123458 Watermelon         Food     8.73

This SQL query returns the most expensive item from each type:
SELECT type, MAX(price) FROM products GROUP BY type

Clothing $34.97
Food     $8.73
Music    $21.55
Toy      $89.95

I also want to get the fields id and name that belong to the above max price, for each row.
What SQL query will return a table like this?

Id     name            type      price
123455 Blouse          Clothing  34.97
123458 Watermelon      Food      8.73
123457 Country Tunes   Music     21.55
123453 Playstation     Toy       89.95

Answer 1

This is the greatest-n-per-group problem that comes up frequently. My usual way of solving it is logically equivalent to the answer given by @Martin Smith, but does not use a subquery:

SELECT T1.Id, T1.name, T1.type, T1.price 
FROM Table T1
LEFT OUTER JOIN Table T2
  ON (T1.type = T2.type AND T1.price < T2.price)
WHERE T2.price IS NULL;

My solution and all others given on this thread so far have a chance of producing multiple rows per value of type, if more than one product shares the same type and both have an equal price that is the max. There are ways to resolve this and break the tie, but you need to tell us which product “wins” in case like that.

You need some other attribute that is guaranteed to be unique over all rows, at least for rows with the same type. For example, if the product with the greater Id value should win, you can resolve the tie this way:

SELECT T1.Id, T1.name, T1.type, T1.price 
FROM Table T1
LEFT OUTER JOIN Table T2
  ON (T1.type = T2.type AND (T1.price < T2.price
       OR T1.price = T2.price AND T1.Id < T2.Id))
WHERE T2.price IS NULL;