Question

尝试获取最新活动，使用不同的user_id，按“最新”（订单“id desc”或“created_at desc”）排序：

SELECT DISTINCT ON (id, user_id) id,user_id 
FROM "activities" 
GROUP BY id,user_id 
ORDER BY id desc LIMIT 10

但是使用以下user_ids返回10个结果：

2863, 2863, 2863, 2863, 2863, 2863, 2863, 2863, 2863, 2615

显然，这并不明显。我如何得到它，以便它有一个独特的（分组的）user_id？

编辑：也许我解释得很糟糕，我道歉。真的很简单，我想要来自唯一user_id的最新活动。根据最新情况，我将更喜欢create_at，但是id也适用。我只想要每个用户一个，没有重复。

再次编辑：意识到第一句确实是在抛弃回答者。很抱歉让所有人都离开那里，我想要的非常简单：/。

Answer 1

试试这个：

select user_id, array_agg(id) as ids
from activities
group by user_id
having count(*) > 1;

array_agg()函数以逗号分隔的列表的形式返回组的所有值。

编辑：

从评论中可能只是想要这个：

select distinct user_id
from activities;

或者可能是这样，它只显示那些具有重复项的用户ID：

select user_id
from activities
group by user_id
having count(*) > 1;

Answer 2

我不确定你想要什么，但试试这个：

SELECT id, user_id
FROM (
   SELECT id
        , user_id
        , ROW_NUMBER () OVER (PARTITION BY id
                              ORDER BY user_id DESC) as rn
   FROM "activities"
   ) as xx
WHERE rn = 1
ORDER BY id desc 
LIMIT 10

ROW_NUMBER OLAP函数将在id。

的值内分配计数器

Answer 3

RE：我想要来自唯一user_id的最新活动

SELECT DISTINCT ON (user_id) user_id, created_at
FROM "activities" 
ORDER BY user_id, created_at desc

RE：那我怎么得到身份证？我还需要ids

重复使用上面的查询：

with latest_activities as
(
  SELECT DISTINCT ON (user_id) user_id, created_at
  FROM "activities" 
  ORDER BY user_id, created_at desc
)
select l.user_id, l.created_at, array_agg(a.id) as ids
from latest_activities l
join activities a using(user_id)
group by l.user_id, l.created_at

Answer 4

如果你有一些最新的关系活动，例如

user_id created_at      id
john    july 4, 2010    1
john    july 4, 2010    2
john    july 12, 2010   3        -- ties with id# 4
john    july 12, 2010   4        -- ties with id# 3
john    july 5, 2010    5
paul    july 13, 2010   6
paul    july 12, 2010   7

使用dense_rank：

with latest_activities as
(
    SELECT user_id, created_at, id,
        dense_rank() 
        over(partition by user_id order by created_at desc) as the_ranking
    FROM activities     
)
select * 
from latest_activities
where the_ranking = 1
order by user_id, id;

上面的查询将显示：

user_id created_at      id 
john    july 12, 2010   3
john    july 12, 2010   4
paul    july 13, 2010   6

如果您希望每个用户的多个ID仅出现在一行上，请使用 ~~group_concat~~ array_agg，然后它将显示：

user_id created_at      ids
john    july 12, 2010   3, 4
paul    july 13, 2010   6

ARRAY_AGG：

with latest_activities as
(
    SELECT user_id, created_at, id,
        dense_rank() 
        over(partition by user_id order by created_at desc) as the_ranking
    FROM activities     
)
select user_id, created_at, array_agg(id order by id) as ids
from latest_activities
where the_ranking = 1
group by user_id, created_at

请注意，上面的查询也在处理没有关系的数据

实时测试：http://www.sqlfiddle.com/#!12/c48ec/8

无法让DISTINCT + GROUP与Postgres一起使用两个值

4 个答案:

编辑：