我下面有一个示例表,其中显示了票证编号,票证打开时间和票证关闭时间。
TKTNUM OPEN_DATE CLOSE_DATE
1234 12-Mar-19 08:36 14-Mar-19 08:36
1235 13-Mar-19 08:36 15-Mar-19 08:36
1236 14-Mar-19 08:36 16-Mar-19 08:36
1237 15-Mar-19 08:36
1238 16-Mar-19 08:36
1239 17-Mar-19 08:36
1240 18-Mar-19 08:36 20-Mar-19 08:36
1241 19-Mar-19 08:36 20-Mar-19 08:36
1242 20-Mar-19 08:36 21-Mar-19 08:36
我需要计算给定日期的开/关票数量...
DATE OPEN CLOSED
12-Mar-19 08:36 1 0
13-Mar-19 08:36 2 0
14-Mar-19 08:36 2 1
15-Mar-19 08:36 2 2
16-Mar-19 08:36 2 3
17-Mar-19 08:36 3 3
18-Mar-19 08:36 4 3
19-Mar-19 08:36 5 3
20-Mar-19 08:36 4 5
任何帮助将不胜感激。谢谢
在示例job_history表上使用了以下查询(c / o Tejash)
EMPLOYEE_ID START_DATE END_DATE JOB_ID DEPARTMENT_ID
----------- -------------------- -------------------- ---------- -------------
200 17/SEP/1995 00:00:00 17/JUN/2001 00:00:00 AD_ASST 90
101 21/SEP/1997 00:00:00 27/OCT/2001 00:00:00 AC_ACCOUNT 110
102 13/JAN/2001 00:00:00 24/JUL/2006 00:00:00 IT_PROG 60
101 28/OCT/2001 00:00:00 15/MAR/2005 00:00:00 AC_MGR 110
200 01/JUL/2002 00:00:00 31/DEC/2006 00:00:00 AC_ACCOUNT 90
201 17/FEB/2004 00:00:00 19/DEC/2007 00:00:00 MK_REP 20
114 24/MAR/2006 00:00:00 31/DEC/2007 00:00:00 ST_CLERK 50
176 24/MAR/2006 00:00:00 31/DEC/2006 00:00:00 SA_REP 80
176 01/JAN/2007 00:00:00 31/DEC/2007 00:00:00 SA_MAN 80
122 01/JAN/2007 00:00:00 31/DEC/2007 00:00:00 ST_CLERK 50
With dates(dt)
As (Select mindt + level - 1 from
(Select min(start_date) mindt, max(end_date) maxdt from job_history)
Connect by level <= maxdt - mindt + 1)
Select dt,
sum(case when dt between start_date and coalesce(end_date,dt) then 1 end) as startdate,
Sum(case when dt >= end_date then 1 end) as enddate
From dates cross join job_history
Group by dt
Order by dt desc
2001年6月17日,查询给出了
DT STARTDATE ENDDATE
-------------------- ---------- ----------
31/DEC/2007 00:00:00 3 10
<SNIPPED>
17/JUN/2001 00:00:00 3 1
代替
DT STARTDATE ENDDATE
-------------------- ---------- ----------
31/DEC/2007 00:00:00 3 10
<SNIPPED>
17/JUN/2001 00:00:00 2 1
试图编辑查询,现在它给了我
DT STARTDATE ENDDATE
-------------------- ---------- ----------
31/DEC/2007 00:00:00 <<< 10
<snipped>
18/JUN/2001 00:00:00 2 1
17/JUN/2001 00:00:00 2 <<< 1
16/JUN/2001 00:00:00 3 1
答案 0 :(得分:1)
您可以取消透视和聚合:
select dte, sum(is_open) as num_opens, sum(is_close) as num_closes
from ((select open_date as dte, 1 as is_open, 0 as is_close
from t
) union all
(select close_date, 0 as is_open, 1 as is_close
from t
)
) t
group by dte
order by dte;
注意:截断日期可能是一个好主意,因此它没有时间部分:
select trunc(dte), sum(is_open) as num_opens, sum(is_close) as num_closes
from ((select open_date as dte, 1 as is_open, 0 as is_close
from t
) union all
(select close_date, 0 as is_open, 1 as is_close
from t
)
) t
where dte is not null
group by trunc(dte)
order by trunc(dte);
在Oracle 12C中,您可以为此使用横向join
:
select trunc(dte), sum(is_open), sum(is_close)
from t cross join lateral
(select t.open_date as dte, 1 as is_open, 0 as is_close from dual union all
select t.close_date, 0 as is_open, 1 as is_close from dual
) t
group by trunc(dte)
order by trunc(dte);
答案 1 :(得分:1)
您可以将日期用作总天数的cte,然后使用以下相同的表格再次将其加入:
With dates(dt)
As
(
Select mindt + level - 1 from
(Select min(open_date) mindt, max(open_dt) maxdt from your_table)
Connect by level <= maxdt - mindt + 1
)
Select dt,
sum(case when dt between open_date and coalesce(close_date,dt) then 1 end) as open,
Sum(case when dt >= close_date then 1 end) as closed
From dates cross join your_table
Group by dt;
干杯!