Sql 按组对行进行排序的查询

我正在使用ApacheDerby10.10

我有一份参与者名单，并希望计算他们在本国的排名，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：

这就是我尝试过的：

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，OVER语句不接受任何参数

---

有人有办法获得国家排名吗？

您只需要添加一个按国家划分的分区，这将为您提供所需的内容

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;

---

考虑使用相关聚合计数子查询的非windows函数SQL查询。由于组列Country.name与排名标准Participant.points不在同一个表中，因此我们需要在子查询中运行相同的联接，但重命名表别名以正确比较内部查询和外部查询

当然，在一个完美的世界里，情况就是这样，但我们现在必须考虑平局。因此，将使用另一个非常类似的tie breaker子查询添加到第一个子查询中。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的字母顺序排列

SQL

在线演示

SQL Fiddle演示：

解释

---

可以使用一个简单的ANSI-SQL子选择来完成相同的工作，计算同一国家/地区中分数较低或分数相同且名称字母顺序不高的参与者的记录数。

谢谢JVM，在Apache Derby中按分区可以工作吗？我试过了，但它说我在第x行第I列遇到了分区，我对ApacheDerby不太熟悉，但我在google上搜索了一下，发现它们确实存在于ApacheDerby中。不幸的是，这是一个链接，那个wiki页面描述了构建此类功能的设计方案，并不是所有的功能最终都得到了构建。向您致敬！你更擅长打平手！

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;

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;

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;