Postgresql OVER和GROUP

Question

我有分组问题

我有一张桌子(其中有很多东西,但不相关)看起来像:

id user
0 1
1 1
2 1
3 2
4 2
5 2
6 1
7 1

我正在尝试获得以下值:

user start end
1 0 2
2 3 5
1 6 7

基本上,我需要用户的第一次和最后一次出现,而不是搞乱订单.我知道我需要使用OVER(PARTITION BY ...),但我从未使用它,也不确定如何构建此查询.如果我"按用户分区",它会忽略顺序.如果我"按id,用户分区"它再次返回错误.

我试过的例子(甚至没有尝试得到我需要的东西,但是一个中点,告诉我一旦我弄清楚"结束"部分怎么做):

SELECT user, count(user) over (partition by user):
user count
1 5
1 5
1 5
2 3
2 3
2 3
1 5
1 5

SELECT user, count(user) over (partition by id, user):
user count
1 1
1 1
1 1
2 1
2 1
2 1
1 1
1 1

Answer 1

使用行号差异方法将具有相同用户的连续行分类到一个组中,并在遇到新用户时重新开始新组.此后,用于group by获取每个组的开始和结束.

SELECT USER,MIN(ID) AS START,MAX(ID) AS END
FROM (SELECT user,id, row_number() over(order by id) 
                     - row_number() over (partition by user order by id) as grp
      FROM tablename
      ) T
GROUP BY USER,GRP

Answer 2

要开始，结束 id，请使用：

SELECT user, min(id) over (partition by user) "start", max(id) over (partition by user) "end"
from table_name;

更新 我的答案是基于错误的谓词，所以是错误的。为了提供正确的而不是重复的@vkp，我做了这个可怕的构造：

create table so74 as
select * from (values (0, 1), (1, 1), (2, 1), (3, 2), (4, 2), (5, 2), (6, 1), (7, 1)) t(id, u); 

with d as (
    with c as (
        with b as (
            select 
                *
                , case when lag(u) over (order by id) <> u or id = min(id) over() then id end min
                , case when lead(u) over (order by id) <> u or id=max(id) over () then id end max 
            from so74
            )
        select  u, min,max 
        from b 
        where coalesce(min,max) is not null
    ) 
    select u,min,lead(max) over () max  
    from c
) 
select * 
from d 
where coalesce(min,max) is not null
;

 u | min | max
---+-----+-----
 1 |   0 |   2
 2 |   3 |   5
 1 |   6 |   7
(3 rows)

Time: 0.456 ms