Postgres 聚合
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【中文标题】Postgres 聚合【英文标题】:Postgres Aggregations 【发布时间】:2020-12-10 10:52:02 【问题描述】:我有一个如下的 postgres 表
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 |
试图实现的结果表如下:
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"
【问题讨论】:
您还没有真正问过问题 - 您当前的输出是什么,您需要什么帮助? 我需要一个查询来帮助我实现第二个表 【参考方案1】: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';
【讨论】:
【参考方案2】:我将此视为“我想要所有登录事件。如果那是下一个事件,我想要下一次注销”。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';
【讨论】:
谢谢!这行得通!选择此作为正确答案,因为这是我尝试的第一个答案【参考方案3】:这听起来像是一个孤岛问题。我会推荐一个登录窗口计数来构建组,然后聚合:
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
【讨论】:
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