LIMIT 基于 COUNT 的 DISTINCT `foreign_key` 到目前为止

Question

我有 2 个表“站点”和“卡车”的联合（按distance排序）。 记录集如下所示：

我需要获取所有行，直到达到特定 (n) 数量的唯一 company_id，从第 1 行开始。

所以，如果我得到如下内容：

然后我可以做一个简单的查询，例如：

SELECT * FROM union_recordset where distinct_company_id_count_so_far < (3 + 1);
-- where n = 3

并获得所需的结果：

Answer 1

如果您的数据库支持count(distinct)作为窗口函数：

select ur.*,
       count(distinct company_id) over (order by distance) as cnt
from union_recordset ur
order by distance;

如果没有，您可以计算第一次出现：

select ur.*,
       sum(case when seqnum = 1 then 1 else 0 end) over (order by distance) as cnt
from (select ur.*,
             row_number() over (partition by company_id order by distance) as seqnum
      from union_recordset ur
     ) ur
order by distance;

在 Postgres 中， sum()可以简化为：

       sum( (seqnum = 1)::int ) over (order by distance) as cnt

然后要获得前三家公司的数字，您需要：

select ur.*
from (select ur.*,
             sum( (seqnum = 1)::int ) over (order by distance) as cnt
      from (select ur.*,
                   row_number() over (partition by company_id order by distance) as seqnum
            from union_recordset ur
           ) ur
     ) ur
where cnt <= 3
order by distance;

Answer 2

您可以选择您的公司，然后将其与自己再次连接以获取其他数据：

select ur.* from union_recordset ur join
  (select distinct company_id from union_recordset order by distance limit 3) ur_d
  on (ur.company_id = ur_d.company_id)

注意：PostgreSQL 8.1 之后支持Limit命令。