Question

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，但每天有数千条路线。我试图找出每个位置，并且v_type大多数车辆在路上的时间（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。

基本思路是，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，在任何时间点找到道路上的车辆数量。

这是我到目前为止所做的，但是它没有完全正确地工作，并且很慢。

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))

Answer 1

我会将此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，你开始沿着这条道路前进。

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1 个答案: