查询以对组中的行进行排名

Question

我正在使用 Apache Derby 10.10。

我有一份参与者名单，想计算他们在他们国家的排名，如下所示：

|        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 |

架构如下所示 (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)
);

这是我试过的：

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;

但是根据 apache derby doco，OVER() 语句不接受任何参数。

有没有人有办法达到国家排名？

Answer 1

您只需添加一个按国家/地区划分的分区，这应该可以满足您的需求。

SELECT
Country.name,
Participant.name,
Participant.points,
ROW_NUMBER() OVER(PARTITION BY country 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;

Answer 2

考虑使用相关聚合计数子查询的非windows 函数SQL 查询。因为组列 (Country.name) 与排名标准 (Participant.points) 不在同一 table 中，我们需要在子查询中运行相同的连接，但重命名 table 别名以正确比较内部和外部查询。

现在当然，在一个完美的世界中就是这样，但我们现在必须考虑平分。因此，另一个非常相似的子查询（用于决胜局）被用于添加到第一个子查询。第二个嵌套查询匹配内部和外部查询的 Country.name 和 Participant.points 但按 Participant.name.

的字母顺序排名

SELECT
    Country.name AS Country,
    Participant.name AS Participant,
    Participant.points,
    (SELECT Count(*) + 1
       FROM Country subC
      INNER JOIN Team subT
              ON subC.id = subT.country_id
      INNER JOIN Participant subP
              ON subT.id = subP.team_id
      WHERE subC.name = Country.name
        AND subP.points < Participant.points) 

     +

    (SELECT Count(*)
       FROM Country subC
      INNER JOIN Team subT
              ON subC.id = subT.country_id
      INNER JOIN Participant subP
              ON subT.id = subP.team_id
      WHERE subC.name = Country.name
        AND subP.points = Participant.points
        AND subP.name < Participant.name)  As country_rank

FROM Country
INNER JOIN Team
        ON Country.id = Team.country_id
INNER JOIN Participant
        ON Team.id = Participant.team_id
ORDER BY Country.name, Participant.points;

Answer 3

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;

在线演示

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

说明

一个简单的 ANSI-SQL subselect 可以用来做同样的工作，计算同一国家中得分较低或得分相同但姓名按字母顺序排列的参与者的记录数没有更高。

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