Question

我正在努力创建一个涉及使用PostgreSQL聚合的SQL查询。请考虑以下表格：

CREATE TABLE thing (
  id INT NOT NULL PRIMARY KEY,
  price NUMERIC(10,2) NOT NULL,
  description VARCHAR(255) NOT NULL,
  url VARCHAR(255) NOT NULL,
  location_id INT NOT NULL REFERENCES location(id)
)

CREATE TABLE location (
  id INT NOT NULL PRIMARY KEY,
  type INT NOT NULL,
  name VARCHAR(255) NOT NULL
)

现在，我想获得每个位置的所有记录，其中location.type = xxx具有最低价格。

类似的东西：

SELECT min(price) FROM thing
INNER JOIN location ON (thing.location_id = location.id)
WHERE type = xxx
GROUP BY location_id

这将列出xxx类型的每个位置的最低价格，但是如何从表格中获取这些列的行（或它们的主键）？

Answer 1

使用此PostgreSQL扩展程序：

SELECT  DISTINCT ON (location.id) thing.*
FROM    location
JOIN    thing
ON      thing.location_id = location_id
WHERE   type = 1
ORDER BY
        location.id ASC, price ASC

这将仅选择每个location.id的第一行。

由于您的行按location.id排序，然后按price排序，因此这将是最低价格的行。

在新的PostgreSQL 8.4中，您还可以使用窗口函数：

SELECT  *
FROM    (
        SELECT  thing.*, ROW_NUMBER() OVER (PARTITION BY location_id ORDER BY price) AS rn
        FROM    location
        JOIN    thing
        ON      thing.location_id = location_id
        WHERE   type = 1
        ) q
WHERE   rn = 1

Answer 2

也许使用子查询

SELECT t.id,t.description,t.price FROM 
  ( SELECT location_id, min(price) FROM thing
      INNER JOIN location ON (thing.location_id = location.id)
      WHERE type = xxx
      GROUP BY location_id
   ) AS lowest
   INNER JOIN thing AS t
     ON t. location_id  = lowest.location_id;

Answer 3

我是SQL Server的人，但以下内容应该符合SQL-92并且应该可以工作：

select th.*
 from thing th
  inner join (select lo.id, min(th.price) minPrice
               from location lo
                inner join thing th
                 on th.location_id = lo.id
               where lo.type = xxx
               group by lo.id) minSet
   on minSet.id = th.location_id
    and th.price = minSet.minPrice

另请注意，我没有设置表格进行测试，因此可能会有一两个错字。

虽然它确实有效但确实看起来很尴尬。如果Postgres有类似SQL的排名函数，它们会让它变得更简单。

Answer 4

尝试此查询

select thing.id,thing.description,thing.url,low.location_id,low.minimum from
(select thing.location_id,min(price) as minimum from thing
 join location on thing.location_id=location.id
  where location.type = 13 group by location_id) as low
   inner join thing on thing.location_id = low.location_id

使用PostgreSQL获取聚合，按SQL查询分组的主键

4 个答案: