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sqlderbyrankrow-numbergroupwise-maximum

Query to rank rows in groups


I'm using Apache Derby 10.10.

I have a list of participants and would like to calculate their rank in their country, like this:

|        Country |         Participant | Points | country_rank |
|----------------|---------------------|--------|--------------|
|      Australia |      Bridget Ciriac |      1 |            1 |
|      Australia |     Austin Bjorklun |      4 |            2 |
|      Australia |        Carrol Motto |      7 |            3 |
|      Australia |     Valeria Seligma |      8 |            4 |
|      Australia |     Desmond Miyamot |     27 |            5 |
|      Australia |      Maryjane Digma |     33 |            6 |
|      Australia |       Kena Elmendor |     38 |            7 |
|      Australia |         Emmie Hicke |     39 |            8 |
|      Australia |        Kaitlyn Mund |     50 |            9 |
|      Australia |    Alisia Vitaglian |     65 |           10 |
|      Australia |          Anika Bulo |     65 |           11 |
|             UK |          Angle Ifil |      2 |            1 |
|             UK |     Demetrius Buelo |     12 |            2 |
|             UK |      Ermelinda Mell |     12 |            3 |
|             UK |         Adeline Pee |     21 |            4 |
|             UK |     Alvera Cangelos |     23 |            5 |
|             UK |   Keshia Mccalliste |     23 |            6 |
|             UK |        Alayna Rashi |     24 |            7 |
|             UK |    Malinda Mcfarlan |     25 |            8 |
|  United States |     Gricelda Quirog |      3 |            1 |
|  United States |      Carmina Britto |      5 |            2 |
|  United States |         Noemi Blase |      6 |            3 |
|  United States |        Britta Swayn |      8 |            4 |
|  United States |        An Heidelber |     12 |            5 |
|  United States |        Maris Padill |     21 |            6 |
|  United States |     Rachele Italian |     21 |            7 |
|  United States |   Jacquiline Speake |     28 |            8 |
|  United States |      Hipolito Elami |     45 |            9 |
|  United States |          Earl Sayle |     65 |           10 |
|  United States |       Georgeann Ves |     66 |           11 |
|  United States |       Conchit Salli |     77 |           12 |

The schema looks like this (sqlfiddle):

create table Country(
  id INTEGER NOT NULL GENERATED ALWAYS AS IDENTITY,
  name varchar(255),
  PRIMARY KEY (id)
);

create table Team(
  id INTEGER NOT NULL GENERATED ALWAYS AS IDENTITY,
  country_id int not null,
  PRIMARY KEY (id),
  FOREIGN KEY (country_id) REFERENCES Country(id)
);

create table Participant(
  id INTEGER NOT NULL GENERATED ALWAYS AS IDENTITY,
  team_id int not null,
  name varchar(100),
  points int,
  PRIMARY KEY (id),
  FOREIGN KEY (team_id) REFERENCES Team(id)
);

This is what I have tried:

select
    Country.name,
    Participant.name,
    Participant.points,
    ROW_NUMBER() OVER(order by Country.name, Participant.points) as country_rank
from Country
join Team
    on Country.id = Team.country_id
join Participant
    on Team.id = Participant.team_id;

But according to the apache derby doco, the OVER() statement doesn't take any arguments.

Does anyone have a way to achieve the country rank?


Solution

  • SQL

    SELECT c.name AS Country,
           p.name AS Participant,
           p.points AS Points,
           (SELECT COUNT(*)
            FROM Participant p2
            JOIN Team t2 ON p2.team_id = t2.id
            WHERE t2.country_id = t.country_id
              AND (p2.points < p.points
                   OR p2.points = p.points AND p2.name <= p.name)) AS country_rank
    FROM Country c
    JOIN Team t ON c.id = t.country_id
    JOIN Participant p ON t.id = p.team_id
    ORDER BY c.name, p.points, p.name;
    

    Online Demo

    SQL Fiddle demo: http://sqlfiddle.com/#!5/f48f8/14

    Explanation

    A simple ANSI-SQL subselect can be used to do the same job, counting the number of records for participants in the same country with a lower score or with the same score and a name that is alphabetically no higher.