Get records with max value for each group of grouped SQL results

Question

The correct solution is:

SELECT o.*
FROM `Persons` o                    # 'o' from 'oldest person in group'
  LEFT JOIN `Persons` b             # 'b' from 'bigger age'
      ON o.Group = b.Group AND o.Age < b.Age
WHERE b.Age is NULL                 # bigger age not found

How it works:

It matches each row from o with all the rows from b having the same value in column Group and a bigger value in column Age. Any row from o not having the maximum value of its group in column Age will match one or more rows from b.

The LEFT JOIN makes it match the oldest person in group (including the persons that are alone in their group) with a row full of NULLs from b ('no biggest age in the group').
Using INNER JOIN makes these rows not matching and they are ignored.

The WHERE clause keeps only the rows having NULLs in the fields extracted from b. They are the oldest persons from each group.

Further readings

This solution and many others are explained in the book SQL Antipatterns: Avoiding the Pitfalls of Database Programming

There's a super-simple way to do this in mysql:

select * 
from (select * from mytable order by `Group`, age desc, Person) x
group by `Group`

This works because in mysql you're allowed to not aggregate non-group-by columns, in which case mysql just returns the first row. The solution is to first order the data such that for each group the row you want is first, then group by the columns you want the value for.

You avoid complicated subqueries that try to find the max() etc, and also the problems of returning multiple rows when there are more than one with the same maximum value (as the other answers would do)

Note: This is a mysql-only solution. All other databases I know will throw an SQL syntax error with the message "non aggregated columns are not listed in the group by clause" or similar. Because this solution uses undocumented behavior, the more cautious may want to include a test to assert that it remains working should a future version of MySQL change this behavior.

Version 5.7 update:

Since version 5.7, the sql-mode setting includes ONLY_FULL_GROUP_BY by default, so to make this work you must not have this option (edit the option file for the server to remove this setting).

You can join against a subquery that pulls the MAX(Group) and Age. This method is portable across most RDBMS.

SELECT t1.*
FROM yourTable t1
INNER JOIN
(
    SELECT `Group`, MAX(Age) AS max_age
    FROM yourTable
    GROUP BY `Group`
) t2
    ON t1.`Group` = t2.`Group` AND t1.Age = t2.max_age;

My simple solution for SQLite (and probably MySQL):

SELECT *, MAX(age) FROM mytable GROUP BY `Group`;

However it doesn't work in PostgreSQL and maybe some other platforms.

In PostgreSQL you can use DISTINCT ON clause:

SELECT DISTINCT ON ("group") * FROM "mytable" ORDER BY "group", "age" DESC;

Not sure if MySQL has row_number function. If so you can use it to get the desired result. On SQL Server you can do something similar to:

CREATE TABLE p
(
 person NVARCHAR(10),
 gp INT,
 age INT
);
GO
INSERT  INTO p
VALUES  ('Bob', 1, 32);
INSERT  INTO p
VALUES  ('Jill', 1, 34);
INSERT  INTO p
VALUES  ('Shawn', 1, 42);
INSERT  INTO p
VALUES  ('Jake', 2, 29);
INSERT  INTO p
VALUES  ('Paul', 2, 36);
INSERT  INTO p
VALUES  ('Laura', 2, 39);
GO

SELECT  t.person, t.gp, t.age
FROM    (
         SELECT *,
                ROW_NUMBER() OVER (PARTITION BY gp ORDER BY age DESC) row
         FROM   p
        ) t
WHERE   t.row = 1;

Using ranking method.

SELECT @rn :=  CASE WHEN @prev_grp <> groupa THEN 1 ELSE @rn+1 END AS rn,  
   @prev_grp :=groupa,
   person,age,groupa  
FROM   users,(SELECT @rn := 0) r        
HAVING rn=1
ORDER  BY groupa,age DESC,person

This sql can be explained as below,

1. select * from users, (select @rn := 0) r order by groupa, age desc, person

2. @prev_grp is null

3. @rn := CASE WHEN @prev_grp <> groupa THEN 1 ELSE @rn+1 END

   this is a three operator expression
   like this, rn = 1 if prev_grp != groupa else rn=rn+1

4. having rn=1 filter out the row you need