I'm trying to write a query in PostgreSQL and I'm getting a little frustrated because it works in other database engines. I need to select the top 5 users from a given joins table like this:
我正在尝试在PostgreSQL中编写一个查询,我有点沮丧,因为它在其他数据库引擎中工作。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;
I need the top 5 users. This code works in sqlite, mysql, etc, yet PostgreSQL refuses to select additional fields that aren't used in aggregate functions. I'm getting the following error:
我需要前5名用户。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
How can I do this in PostgreSQL??
我怎么能在PostgreSQL中做到这一点?
4 个解决方案
#1
7
You could try:
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
#2
2
Assuming that users.id IS a PK, then you can either
假设users.id是PK,那么你也可以
wait for 9.1
等9.1
group by all fields
按所有领域分组
use an aggregate (i.e. max() ) on all fields
在所有字段上使用聚合(即max())
#3
2
One other solution that works is to use all attributes implicitly in GROUP BY
另一个有效的解决方案是在GROUP BY中隐式使用所有属性
Thus following will be final query
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;
If you are using framework like rails then you can implement this easily with Model.column_names function.
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。
#4
0
Just in case of somebody wants ANSI-92 standard solution and doesn't like 'Oracle' way to join tables...
如果有人想要ANSI-92标准解决方案并且不喜欢'Oracle'加入表的方式......
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;
#1
7
You could try:
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
#2
2
Assuming that users.id IS a PK, then you can either
假设users.id是PK,那么你也可以
wait for 9.1
等9.1
group by all fields
按所有领域分组
use an aggregate (i.e. max() ) on all fields
在所有字段上使用聚合(即max())
#3
2
One other solution that works is to use all attributes implicitly in GROUP BY
另一个有效的解决方案是在GROUP BY中隐式使用所有属性
Thus following will be final query
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;
If you are using framework like rails then you can implement this easily with Model.column_names function.
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。
#4
0
Just in case of somebody wants ANSI-92 standard solution and doesn't like 'Oracle' way to join tables...
如果有人想要ANSI-92标准解决方案并且不喜欢'Oracle'加入表的方式......
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;