MySQL: Select N rows, but with only unique values in one column

Question

Given this data set:

ID  Name            City            Birthyear
1   Egon Spengler   New York        1957
2   Mac Taylor      New York        1955
3   Sara         


        

Answer 1

@BlaM

UPDATED just found that its good to use USING instead of ON. it will remove duplicate columns in result.

SELECT P.*, COUNT(*) AS ct
   FROM people P
   JOIN (SELECT City, MIN(Birthyear) AS Birthyear
              FROM people 
              GROUP by City) P2 USING(Birthyear, City)
   GROUP BY P.City
   ORDER BY P.Birthyear ASC 
   LIMIT 10;

ORIGINAL POST

hi, i've tried to use your updated query but i was getting wrong results until i've added extra condition to join (also extra column into join select). transfered to your query, i'am using this:

SELECT P.*, COUNT(*) AS ct
   FROM people P
   JOIN (SELECT City, MIN(Birthyear) AS Birthyear
              FROM people 
              GROUP by City) P2 ON P2.Birthyear = P.Birthyear AND P2.City = P.City
   GROUP BY P.City
   ORDER BY P.Birthyear ASC 
   LIMIT 10;

in theory you should not need last GROUP BY P.City, but i've left it there for now, just in case. will probably remove it later.

Answer 2

This is probably not the most elegant and quickest solution, but it should work. I am looking forward the see the solutions of real database gurus.

select p.* from people p,
(select city, max(age) as mage from people group by city) t
where p.city = t.city and p.age = t.mage
order by p.age desc

Answer 3

Something like that?

SELECT
  Id, Name, City, Birthyear
FROM
  TheTable
WHERE
  Id IN (SELECT TOP 1 Id FROM TheTable i WHERE i.City = TheTable.City ORDER BY Birthyear)

Answer 4

Probably not the most elegant of solutions, and the performance of IN may suffer on larger tables.

The nested query gets the minimum Birthyear for each city. Only records who have this Birthyear are matched in the outer query. Ordering by age then limiting to 3 results gets you the 3 oldest people who are also the oldest in their city (Egon Spengler drops out..)

SELECT Name, City, Birthyear, COUNT(*) AS ct
FROM table
WHERE Birthyear IN (SELECT MIN(Birthyear)
               FROM table
               GROUP by City)
GROUP BY City
ORDER BY Birthyear DESC LIMIT 3;

+-----------------+-------------+------+----+
| name            | city        | year | ct |
+-----------------+-------------+------+----+
| Henry Jones     | Chicago     | 1899 | 1  |
| Mac Taylor      | New York    | 1955 | 1  |
| Sarah Connor    | Los Angeles | 1959 | 1  |
+-----------------+-------------+------+----+

Edit - added GROUP BY City to outer query, as people with same birth years would return multiple values. Grouping on the outer query ensures that only one result will be returned per city, if more than one person has that minimum Birthyear. The ct column will show if more than one person exists in the city with that Birthyear

Answer 5

Not pretty but should work also with multiple people with the same dob:

Test data:

select id, name, city, dob 
into people
from
(select 1 id,'Egon Spengler' name, 'New York' city , 1957 dob
union all select 2, 'Mac Taylor','New York', 1955
union all select 3, 'Sarah Connor','Los Angeles', 1959
union all select 4, 'Jean-Luc Picard','La Barre', 2305
union all select 5, 'Ellen Ripley','Nostromo', 2092
union all select 6, 'James T. Kirk','Riverside', 2233
union all select 7, 'Henry Jones','Chicago', 1899
union all select 8, 'Blah','New York', 1955) a

Query:

select 
    * 
from 
    people p
    left join people p1
    ON 
        p.city = p1.city
        and (p.dob > p1.dob and p.id <> p1.id)
        or (p.dob = p1.dob and p.id > p1.id)
where
    p1.id is null
order by 
    p.dob