我需要根据连续的工作日期范围对一天的行进行分组和汇总。
表出勤率定义：
row_no NUMBER (*,0) NOT NULL, -- row number - generated from a sequence
worker_id NUMBER NOT NULL, -- Attendance worker id
date1 DATE DEFAULT SYSDATE NOT NULL, -- Attendance Date/time
type1 NUMBER(3,0) NOT NULL, -- Attendance type: 0-Enter, 1-Exit
worker_id date1 type1
2 13/06/2016-09:00 0
3 13/06/2016-12:10 0
2 13/06/2016-13:20 1
2 13/06/2016-15:00 0
2 13/06/2016-17:00 1
3 13/06/2016-18:45 1
2 13/06/2016-19:00 0
如果报告在22:00运行
，则结果worker_id date1 fr_hour to_hour hours
2 13/06/2016 09:00 13:20 4:20
2 13/06/2016 15:00 17:00 2:00
2 13/06/2016 19:00 22:00 3:00
3 13/06/2016 12:10 18:45 6:35
答案 0 :(得分：1)
在内部查询中，我们为每一行获取同一工作人员的下一行（LEAD 1）中的date1，type1，而不是仅过滤我们需要的内容：
SELECT worker_id,
TRUNC (date1) AS date1,
TO_CHAR (date1, 'HH24:MI') fr_hour,
TO_CHAR (date2, 'HH24:MI') to_hour,
TRUNC ( (date2 - date1) * 24) || ':' ||
TO_CHAR (TRUNC ( (date2 - date1) * 24 * 60) - TRUNC ( (date2 - date1) * 24) * 60, '00') hours
FROM (SELECT a.*,
LEAD (a.date1, 1) OVER (PARTITION BY worker_id ORDER BY date1) date2,
LEAD (a.type1, 1) OVER (PARTITION BY worker_id ORDER BY date1) type2
FROM testtemp a)
WHERE type1 = 0
AND type2 = 1
AND TRUNC (date1) = TRUNC (date2)
答案 1 :(得分：0)
连续一段时间从较早的一天开始会使其复杂化。您可以计算所有日期的所有范围等，回到开始时间 - 假设您没有归档旧记录，并且数据中任何工作人员的第一个条目不是签出 - 然后在您感兴趣的日期完成所有工作过滤后。或者您只能查看该日期的数据，并查看工作人员的记录是否以签到或退房开始。
我为第四名工人添加了记录：
WORKER_ID DATE1 TYPE1
---------- ---------------- ----------
4 2016-06-12 19:00 0
4 2016-06-13 03:00 1
2 2016-06-13 09:00 0
3 2016-06-13 12:10 0
4 2016-06-13 13:00 0
2 2016-06-13 13:20 1
4 2016-06-13 14:30 1
2 2016-06-13 15:00 0
2 2016-06-13 17:00 1
3 2016-06-13 18:45 1
4 2016-06-13 19:00 0
2 2016-06-13 19:00 0
您可以使用分析函数计算每个条目的行号，还可以找到当天每个工作人员的第一个type1值;这也有效地支持了作为单独列的进出时间：
select worker_id, trunc(date1) as date1, type1,
case when type1 = 0 then date1 end as time_in,
case when type1 = 1 then date1 end as time_out,
row_number() over (partition by worker_id, trunc(date1), type1 order by date1) as rn,
min(type1) keep (dense_rank first order by date1) over (partition by worker_id, trunc(date1)) as open_start,
max(type1) keep (dense_rank last order by date1) over (partition by worker_id, trunc(date1)) as open_end,
row_number() over (partition by worker_id, trunc(date1), type1 order by date1)
- case when type1 = 1 then min(type1) keep (dense_rank first order by date1)
over (partition by worker_id, trunc(date1)) else 0 end as grp
from attendance
where date1 >= date '2016-06-13' and date1 < date '2016-06-14'
order by worker_id, attendance.date1;
WORKER_ID DATE1 TYPE1 TIME_IN TIME_OUT RN OPEN_START OPEN_END GRP
---------- ---------------- ---------- ---------------- ---------------- ---------- ---------- ---------- ----------
2 2016-06-13 00:00 0 2016-06-13 09:00 1 0 0 1
2 2016-06-13 00:00 1 2016-06-13 13:20 1 0 0 1
2 2016-06-13 00:00 0 2016-06-13 15:00 2 0 0 2
2 2016-06-13 00:00 1 2016-06-13 17:00 2 0 0 2
2 2016-06-13 00:00 0 2016-06-13 19:00 3 0 0 3
3 2016-06-13 00:00 0 2016-06-13 12:10 1 0 1 1
3 2016-06-13 00:00 1 2016-06-13 18:45 1 0 1 1
4 2016-06-13 00:00 1 2016-06-13 03:00 1 1 0 0
4 2016-06-13 00:00 0 2016-06-13 13:00 1 1 0 1
4 2016-06-13 00:00 1 2016-06-13 14:30 2 1 0 1
4 2016-06-13 00:00 0 2016-06-13 19:00 2 1 0 2
rn列是原始（天真）尝试将记录分组在一起，但对于工作人员4来说是不合时宜的。如果第一张唱片是退房，则open_start可以解决。然后可以从rn中减去得到的值（0或1），以获得更有用的分组标记，我称之为grp。
然后，您可以将其用作内联视图或CTE，并汇总每个组的时间输入/输出记录，添加nvl()或coalesce()以放入缺少的午夜开始或10 pm-end值：
select worker_id,
date1 as date1,
nvl(min(time_in), date1) as fr_hour,
nvl(max(time_out), date1 + 22/24) as to_hour,
date1 + (nvl(max(time_out), date1 + 22/24) - date1)
- (nvl(min(time_in), date1) - date1) as hours
from (
select worker_id,
trunc(date1) as date1,
case when type1 = 0 then date1 end as time_in,
case when type1 = 1 then date1 end as time_out,
row_number() over (partition by worker_id, trunc(date1), type1 order by date1)
- case when type1 = 1 then min(type1) keep (dense_rank first order by date1)
over (partition by worker_id, trunc(date1)) else 0 end as grp
from attendance
where date1 >= date '2016-06-13' and date1 < date '2016-06-14'
)
group by worker_id, date1, grp
order by worker_id, date1, grp;
WORKER_ID DATE1 FR_HOUR TO_HOUR HOURS
---------- ---------------- ---------------- ---------------- ----------------
2 2016-06-13 00:00 2016-06-13 09:00 2016-06-13 13:20 2016-06-13 04:20
2 2016-06-13 00:00 2016-06-13 15:00 2016-06-13 17:00 2016-06-13 02:00
2 2016-06-13 00:00 2016-06-13 19:00 2016-06-13 22:00 2016-06-13 03:00
3 2016-06-13 00:00 2016-06-13 12:10 2016-06-13 18:45 2016-06-13 06:35
4 2016-06-13 00:00 2016-06-13 00:00 2016-06-13 03:00 2016-06-13 03:00
4 2016-06-13 00:00 2016-06-13 13:00 2016-06-13 14:30 2016-06-13 01:30
4 2016-06-13 00:00 2016-06-13 19:00 2016-06-13 22:00 2016-06-13 03:00
“小时”值操纵日期和时间输入/输出值以显示另一个时间;但它实际上是经过的时间。
最后，您可以格式化列以删除您不感兴趣的位：
select worker_id,
to_char(date1, 'DD/MM/YYYY') as date1,
to_char(nvl(min(time_in), date1), 'HH24:MI') as fr_hour,
to_char(nvl(max(time_out), date1 + 22/24), 'HH24:MI') as to_hour,
to_char(date1 + (nvl(max(time_out),date1 + 22/24) - date1)
- (nvl(min(time_in), date1) - date1), 'HH24:MI') as hours
from (
select worker_id,
trunc(date1) as date1,
case when type1 = 0 then date1 end as time_in,
case when type1 = 1 then date1 end as time_out,
row_number() over (partition by worker_id, trunc(date1), type1 order by date1)
- case when type1 = 1 then min(type1) keep (dense_rank first order by date1)
over (partition by worker_id, trunc(date1)) else 0 end as grp
from attendance
where date1 >= date '2016-06-13' and date1 < date '2016-06-14'
)
group by worker_id, date1, grp
order by worker_id, date1, grp;
WORKER_ID DATE1 FR_HO TO_HO HOURS
---------- ---------- ----- ----- -----
2 13/06/2016 09:00 13:20 04:20
2 13/06/2016 15:00 17:00 02:00
2 13/06/2016 19:00 22:00 03:00
3 13/06/2016 12:10 18:45 06:35
4 13/06/2016 00:00 03:00 03:00
4 13/06/2016 13:00 14:30 01:30
4 13/06/2016 19:00 22:00 03:00