postgreSQL选择未在聚合函数中使用的其他列

时间:2011-01-21 04:55:02

标签: database postgresql aggregate-functions

我正在尝试在PostgreSQL中编写一个查询,但我有点沮丧,因为它可以在其他数据库引擎中运行。我需要从给定的连接表中选择前5个用户,如下所示:

SELECT users.*, 
       COUNT(deals.id) AS num_deals 
FROM users, deals 
WHERE deals.users_id = users.id 
GROUP BY users.id 
ORDER BY num_deals LIMIT 5;

我需要前5位用户。此代码适用于sqlite,mysql等,但PostgreSQL拒绝选择未在聚合函数中使用的其他字段。我收到以下错误:

PGError: ERROR:  column "users.id" must appear in the GROUP BY clause or be used in an aggregate function

如何在PostgreSQL中执行此操作?

4 个答案:

答案 0 :(得分:7)

你可以尝试:

SELECT users.*, a.num_deals FROM users, (
    SELECT deal.id as dealid, COUNT(deals.id) AS num_deals 
    FROM deals 
    GROUP BY deal.id
) a where users.id = a.dealid
ORDER BY a.num_deals DESC
LIMIT 5

答案 1 :(得分:2)

假设users.id是PK,那么你可以

等待9.1

按所有字段分组

在所有字段上使用聚合(即max())

答案 2 :(得分:2)

另一个有效的解决方案是在GROUP BY中隐式使用所有属性

因此,以下将是最终查询

SELECT users.*, 
       COUNT(deals.id) AS num_deals 
FROM users, deals 
WHERE deals.users_id = users.id 
GROUP BY users.id, users.name, users.attrib1, ..., users.attribN
ORDER BY num_deals LIMIT 5;

如果您使用的是类似rails的框架,那么您可以使用Model.column_names函数轻松实现此功能。

答案 3 :(得分:0)

如果有人想要ANSI-92标准解决方案并且不喜欢'Oracle'加入表格的方式......

SELECT users.*, num_deals
FROM users
JOIN
  (SELECT deals.users_id as users_id, count(deals.users_id) as num_deals
   FROM deals
   GROUP BY deals.id) grouped_user_deals
ON grouped_user_deals.users_id = users.id
ORDER BY num_deals DESC
LIMIT 5;