1

我有一张桌子，可以跟踪车辆进出的时间表。

Table: Schedule
route  location    v_type    out                  in
===================================================================
a      loc1        10      2/14/2013 08:04:00   2/14/2013 10:03:00
b      loc1        11      2/14/2013 08:06:00   2/14/2013 14:20:00
c      loc2        11      2/14/2013 06:22:00   2/14/2013 07:50:00
d      loc1        10      2/14/2013 11:04:00   2/14/2013 10:03:00
e      loc2        10      2/14/2013 08:06:00   2/14/2013 14:20:00
f      loc2        11      2/14/2013 06:22:00   2/14/2013 07:50:00

想象一下，每天有数千条路线。我正在尝试找出每个位置，以及 v_type 最多车辆在路上行驶的时间（或时间窗口）。

期望的结果，例如

location    v_type   time            peak
===========================================
loc1        10       2/14/2013 10:40 110
loc1        11       2/14/2013 10:30 80
loc2        10       2/14/2013 08:05 67
loc2        11       2/14/2013 09:45 107

等等

基本思想是，您可以通过查找已离开的车辆总数并减去当天返回的车辆数量来查找任何时间点上道路上的车辆数量。

这是我到目前为止所拥有的，但它不能完全正确地工作，而且速度很慢。

SELECT s.location,
  s.v_type,
  TO_CHAR(TRUNC(s.out, 'mi') - mod(EXTRACT(minute FROM CAST(s.out AS TIMESTAMP)), 10) / (24 * 60), 'YYYY-MM-DD HH24:MI') AS TIME,
  (SELECT
    (SELECT COUNT(*)
     FROM SCHEDULE s2
     WHERE s2.out BETWEEN TRUNC(s.out) AND (TRUNC(s.out, 'mi') - mod(EXTRACT(minute  FROM CAST(s.out AS TIMESTAMP)), 10) / (24 * 60))
    )                                                          
    -
    (SELECT COUNT(*)
     FROM SCHEDULE s2
     WHERE s2.out BETWEEN TRUNC(s.in) AND (TRUNC(s.in, 'mi') - mod(EXTRACT(minute FROM CAST(s.in AS TIMESTAMP)), 10) / (24 * 60))
    )
 FROM dual
 )
 FROM SCHEDULE s
 GROUP BY s.location, s.v_type,
 (TRUNC(s.out, 'mi') - mod(EXTRACT(minute FROM CAST(s.out AS TIMESTAMP)), 10) / (24 * 60))
4

1 回答 1

0

我会将其视为累积金额之间的差异：

select s.*
from (select s.*, (numout - numin) as onroad,
             row_number() over (partition by loc, vtype order by numout - numin desc) as seqnum
      from (select s.*,
                   (select count(*) from schedule s2 where s2.loc = s.loc and s2.vtype = s.vtype and s2.out <= s.out
                   ) as numout,
                   (select count(*) from schedule s2 where s2.loc = s.loc and s2.vtype = s.vtype and and s2.in <= s.out
                   ) as numin
            from schedule s
           ) s
     ) s
where seq = 1

您也可以使用分析函数执行此操作，但相关子查询可能更容易编写。此外，你开始走这条路。

于 2013-02-14T21:54:36.617 回答