问题描述
我有下表：
1）我的公司表格
id | c_name | c_code | status
----+------------+----------+--------
1 | AAAAAAAAAA | AA1234 | Active
2）我的用户表
id | c_id | u_name | status | emp_id
----+------------+----------+--------+--------
1 | 1 | XXXXXXXX | Active | 1
2 | 1 | YYYYYYYY | Active | 2
3）我的出勤表
id | u_id | swipe_time | status
----+--------+------------------------+--------
1 | 1 | 2020-08-20 16:00:00 | IN
2 | 1 | 2020-08-20 20:00:00 | OUT
3 | 1 | 2020-08-20 21:00:00 | IN
4 | 1 | 2020-08-21 01:00:00 | OUT
5 | 1 | 2020-08-21 16:00:00 | IN
6 | 1 | 2020-08-21 19:00:00 | OUT
我需要按日期u_id分组计算出勤率，如下所示：
注意：查询参数为“从日期”，“到日期”和“公司ID”
u_id | u_name | date | in_time | out_time | hrs
-----+-----------+-------------+----------------------+----------------------+-----
1 | XXXXXXXX | 2020-08-20 | 2020-08-20 16:00:00 | 2020-08-21 01:00:00 | 7
1 | XXXXXXXX | 2020-08-21 | 2020-08-21 16:00:00 | 2020-08-21 19:00:00 | 4
2 | YYYYYYYY | null | null | null | 0
这在Postgresql中可能吗？
解决方法
使用lead窗口函数使其更简单易读。对于进出平衡的出勤事件，此方法可以正常工作，否则出勤时间为空值。这是有道理的，因为该人尚未离开或尚未参加会议或出勤数据已损坏。
select
u.id u_id,u.u_name,t.date_in date,t.t_in in_time,t.t_out out_time,extract('hour' from t.t_out - t.t_in) hrs
from users u
left outer join
(
select u_id,date_trunc('day',swipe_time) date_in,swipe_time t_in,lead(swipe_time,1) over (partition by u_id order by u_id,swipe_time) t_out,status
from attendance
) t
on u.id = t.u_id
where t.status = 'IN';
棘手的部分是将涵盖两天（日历）的一行扩展为两行，并正确分配“第二天”的小时数。
第一部分是获取将IN / OUT对组合为单行的数据透视表。
一种简单（但不是很有效）的方法是：
select ain.u_id,ain.swipe_time as time_in,(select min(aout.swipe_time)
from attendance aout
where aout.u_id = ain.u_id
and aout.status = 'OUT'
and aout.swipe_time > ain.swipe_time) as time_out
from attendance ain
where ain.status = 'IN'
下一步是将一天以上的行分成两行。
这假设您的IN / OUT对永远不会超过两天！
with inout as (
select ain.u_id,(select min(aout.swipe_time)
from attendance aout
where aout.u_id = ain.u_id
and aout.status = 'OUT'
and aout.swipe_time > ain.swipe_time) as time_out
from attendance ain
where ain.status = 'IN'
),expanded as (
select u_id,time_in::date as "date",time_in,time_out
from inout
where time_in::date = time_out::date
union all
select i.u_id,x.time_in::date as date,x.time_in,x.time_out
from inout i
cross join lateral (
select i.u_id,i.time_in,i.time_in::date + 1 as time_out
union all
select i.u_id,i.time_out::date,i.time_out
) x
where i.time_out::date > i.time_in::date
)
select *
from expanded;
以上内容为您的示例数据返回了以下内容：
u_id | date | time_in | time_out
-----+------------+---------------------+--------------------
1 | 2020-08-20 | 2020-08-20 16:00:00 | 2020-08-20 20:00:00
1 | 2020-08-20 | 2020-08-20 21:00:00 | 2020-08-21 00:00:00
1 | 2020-08-21 | 2020-08-21 00:00:00 | 2020-08-21 01:00:00
1 | 2020-08-21 | 2020-08-21 16:00:00 | 2020-08-21 19:00:00
这是如何工作的？
因此，我们首先选择与此部分在同一天开始和结束的所有行：
select u_id,time_out
from inout
where time_in::date = time_out::date
联合的第二部分使用交叉连接将跨越两天的行进行拆分，该交叉连接生成的行的原始时间为原始开始时间和午夜，另一行则从午夜开始至原始结束时间：
select i.u_id,x.time_out
from inout i
cross join lateral (
-- this generates a row for the first of the two days
select i.u_id,i.time_in::date + 1 as time_out
union all
-- this generates the row for the next day
select i.u_id,i.time_out
) x
where i.time_out::date > i.time_in::date
最后，通过按用户和日期对新的“扩展”行进行汇总，然后将其合并到users表中以获取用户名。
with inout as (
select ain.u_id,i.time_out
) x
where i.time_out::date > i.time_in::date
)
select u.id,e."date",min(e.time_in) as time_in,max(e.time_out) as time_out,sum(e.time_out - e.time_in) as duration
from users u
left join expanded e on u.id = e.u_id
group by u.id,e."date"
order by u.id,e."date";
这将导致：
u_id | date | time_in | time_out | duration
-----+------------+---------------------+---------------------+----------------------------------------------
1 | 2020-08-20 | 2020-08-20 16:00:00 | 2020-08-21 00:00:00 | 0 years 0 mons 0 days 7 hours 0 mins 0.0 secs
1 | 2020-08-21 | 2020-08-21 00:00:00 | 2020-08-21 19:00:00 | 0 years 0 mons 0 days 4 hours 0 mins 0.0 secs