Postgres 聚合

Question

我有一个 postgres table 如下

Username	Event	Date
UserA	Log in	02/03/2020 07:06:30
UserA	Log in	02/03/2020 10:15:15
UserA	Log in	02/03/2020 10:17:01
UserA	Log Out	02/03/2020 10:28:55
UserA	Log in	02/07/2019 14:56:15
UserA	Log in	02/08/2019 10:50:34
UserA	Log Out	02/08/2019 10:57:21

尝试实现的结果table如下：

Username	log_in_Event	log_in_Date	log_out_event	log_out_date
UserA	Log in	02/03/2020 07:06:30	null	null
UserA	Log in	02/03/2020 10:15:15	null	null
UserA	Log in	02/03/2020 10:17:01	Log Out	02/03/2020 10:28:55
UserA	Log in	02/07/2019 14:56:15	null	null
UserA	Log in	02/08/2019 10:50:34	Log Out	02/08/2019 10:57:21

我已经尝试过的查询如下：

select * from 
(
select  "User Name" , "Event" , "Date" , "IP Address" 
from log_activities log_in 
where "Event" = 'User Logged In'
)log_in
left join 
(
select  "User Name" , "Event" , "Date" , "IP Address" 
from log_activities log_out 
where "Event" = 'User Logged Out'
)log_out
on
log_in."User Name" = log_out."User Name" 
and TO_DATE(log_in."Date" ,'DD/MM/YYYY') = TO_DATE(log_out."Date" ,'DD/MM/YYYY')
and log_in."Date"  < log_out."Date" 
and log_in."IP Address" = log_out."IP Address"

Answer 1

这看起来像是一个缺口和孤岛问题。我建议使用 window 的登录次数来构建组，然后进行聚合：

select username, 
    'Log in' as log_in_event,
    min(date) as log_in_date,
    max(event) filter(where event = 'Log Out') as log_out_event,
    max(date)  filter(where event = 'Log Out') as log_out_date
from (
    select la.*,
        count(*) 
            filter(where event = 'Log in') 
            over(partition by username order by date) as grp
    from log_activities la
) la
group by username, grp

Answer 2

我将其视为“我想要所有登录事件。如果那是下一个事件，我想要下一个注销”。如果是这样，lead() 似乎是最有用的方法：

select la.username, la.event as login_event, la.date as login_date,
       la.next_event as logout_event, la.next_date as logout_date
from (select la.*
             lead(event) over (partition by username order by date) as next_event,
             lead(date) over (partition by username order by date) as next_date
      from log_activities la
     ) la
where event = 'Log in';

Answer 3

您可以使用lead解析函数和CASE..WHEN如下：

select t.username, 
       t.event as login_event, 
       t.date as login_date,
       case when t.lead_event = 'Log Out' then t.lead_event end as logout_event, 
       case when t.lead_event = 'Log Out' then t.lead_date end as logout_date
from (select t.*,
             lead(event) over (partition by username order by date) as lead_event,
             lead(date) over (partition by username order by date) as lead_date
      from log_activities t
     ) t
where t.event = 'Log in';

Postgres 聚合

Postgres Aggregations

sql

postgresql

datetime

gaps-and-islands