我有一个veh_speed表，其中包含字段vid，date_time，speed，status。我的目标是获得速度大于30的车辆的持续时间（start_date_time和end_date_time）。目前我正在使用PL/SQL生成报告。是否可以使用SQL。如果可以在范围之间获得max_speed，那将会很棒。
我的表格如下：
VID START_DATE_TIME SPEED STATUS
--- ------------------- ----- ------
1 15/01/2014 10:00:05 0 N
1 15/01/2014 10:00:10 10 Y
1 15/01/2014 10:00:15 30 Y
1 15/01/2014 10:00:20 35 Y
1 15/01/2014 10:00:25 45 Y
1 15/01/2014 10:00:27 10 Y
1 15/01/2014 10:00:29 0 Y
1 15/01/2014 10:00:30 20 Y
1 15/01/2014 10:00:35 32 Y
1 15/01/2014 10:00:40 33 Y
1 15/01/2014 10:00:45 35 Y
1 15/01/2014 10:00:50 38 Y
1 15/01/2014 10:00:55 10 Y
我想得到以下输出：
VID START_DATE_TIME END_DATE_TIME MAX_SPEED
--- --------------- ------------- ---------
1 15/01/2014 10:00:15 15/01/2014 10:00:25 45
1 15/01/2014 10:00:35 15/01/2014 10:00:50 38
这是表创建脚本：
CREATE TABLE veh_speed(vid NUMBER(3),
date_time DATE,
speed NUMBER(3),
status CHAR(1));
INSERT ALL
INTO veh_speed VALUES(1, to_date('15/01/2014 10:00:05', 'dd/mm/yyyy hh24:mi:ss'), 0, 'N')
INTO veh_speed VALUES(1, to_date('15/01/2014 10:00:10', 'dd/mm/yyyy hh24:mi:ss'), 10, 'Y')
INTO veh_speed VALUES(1, to_date('15/01/2014 10:00:15', 'dd/mm/yyyy hh24:mi:ss'), 30, 'Y')
INTO veh_speed VALUES(1, to_date('15/01/2014 10:00:20', 'dd/mm/yyyy hh24:mi:ss'), 35, 'Y')
INTO veh_speed VALUES(1, to_date('15/01/2014 10:00:25', 'dd/mm/yyyy hh24:mi:ss'), 45, 'Y')
INTO veh_speed VALUES(1, to_date('15/01/2014 10:00:27', 'dd/mm/yyyy hh24:mi:ss'), 10, 'Y')
INTO veh_speed VALUES(1, to_date('15/01/2014 10:00:29', 'dd/mm/yyyy hh24:mi:ss'), 0, 'Y')
INTO veh_speed VALUES(1, to_date('15/01/2014 10:00:30', 'dd/mm/yyyy hh24:mi:ss'), 20, 'Y')
INTO veh_speed VALUES(1, to_date('15/01/2014 10:00:35', 'dd/mm/yyyy hh24:mi:ss'), 32, 'Y')
INTO veh_speed VALUES(1, to_date('15/01/2014 10:00:40', 'dd/mm/yyyy hh24:mi:ss'), 33, 'Y')
INTO veh_speed VALUES(1, to_date('15/01/2014 10:00:45', 'dd/mm/yyyy hh24:mi:ss'), 35, 'Y')
INTO veh_speed VALUES(1, to_date('15/01/2014 10:00:50', 'dd/mm/yyyy hh24:mi:ss'), 38, 'Y')
INTO veh_speed VALUES(1, to_date('15/01/2014 10:00:55', 'dd/mm/yyyy hh24:mi:ss'), 10, 'Y')
SELECT * FROM dual;
我希望我明白我的问题。
提前致谢。
答案 0 :(得分：5)
您可以使用分析功能将记录分组为速度为30或更高的块：
select vid, date_time, speed, status,
case when speed >= 30 then 30 else 0 end as speed_limit,
row_number() over (partition by vid order by date_time)
- row_number() over (
partition by vid, case when speed >= 30 then 30 else 0 end
order by date_time) as chain
from veh_speed;
VID DATE_TIME SPEED STATUS SPEED_LIMIT CHAIN
---------- ------------------- ---------- ------ ----------- ----------
1 15/01/2014 10:00:05 0 N 0 0
1 15/01/2014 10:00:10 10 Y 0 0
1 15/01/2014 10:00:15 30 Y 30 2
1 15/01/2014 10:00:20 35 Y 30 2
1 15/01/2014 10:00:25 45 Y 30 2
1 15/01/2014 10:00:27 10 Y 0 3
1 15/01/2014 10:00:29 0 Y 0 3
1 15/01/2014 10:00:30 20 Y 0 3
1 15/01/2014 10:00:35 32 Y 30 5
1 15/01/2014 10:00:40 33 Y 30 5
1 15/01/2014 10:00:45 35 Y 30 5
1 15/01/2014 10:00:50 38 Y 30 5
1 15/01/2014 10:00:55 10 Y 0 7
我不能赞成使用两个row_number()调用来生成记录链的技巧，不幸的是，我在某个地方选择了它（可能是here）。 chain的实际值无关紧要，只是它们在每个vid中都是唯一的，并且对于符合条件的连续记录块中的所有记录都是相同的。
您只对“速度限制”为30的相关记录链感兴趣（这可能很容易成为Y / N标志或其他），因此您可以使用它并过滤出那些连锁店的速度不到30;然后使用普通的聚合函数来得到你想要的东西：
select vid,
min(date_time) as start_date_time,
max(date_time) as end_date_time,
max(speed) as max_speed
from (
select vid, date_time, speed, status,
case when speed >= 30 then 30 else 0 end as speed_limit,
row_number() over (partition by vid order by date_time)
- row_number() over (
partition by vid, case when speed >= 30 then 30 else 0 end
order by date_time) as chain
from veh_speed
)
where speed_limit = 30
group by vid, chain
order by vid, start_date_time;
VID START_DATE_TIME END_DATE_TIME MAX_SPEED
---------- ------------------- ------------------- ----------
1 15/01/2014 10:00:15 15/01/2014 10:00:25 45
1 15/01/2014 10:00:35 15/01/2014 10:00:50 38
答案 1 :(得分：2)
这个问题众所周知，作为群组的开头，你可以google这个。 通用方法是 a）确定不同行满足标准的标准 c）按正确的顺序对它们进行排序 d）为每个时期制作一个组列，以便及时拆分 e）将它们分组。
就像特殊情况的例子一样：
SQL> select vid, min(date_time) start_time, max(date_time) end_time, max(speed) max_speed
2 from (
3 select vid, date_time,
4 date_time - (row_number() over(partition by vid order by date_time))*speed_sign*5/24/3600 group_time, speed_sign, speed
5 from (
6 select vid, date_time, decode(sign(speed-30),0,1,sign(speed-30)) speed_sign , speed
7 from veh_speed order by date_time
8 )) where speed_sign > 0
9 group by vid, group_time
10 /
VID START_TIME END_TIME MAX_SPEED
1 15.01.2014 10:00:15 15.01.2014 10:00:25 45
1 15.01.2014 10:00:35 15.01.2014 10:00:50 38
答案 2 :(得分：1)
我使用子请求来分组（但我想这不像Alex的解释那么清楚）：
select z.vid, min(z.date_time) start_time, z.end_time, max(z.speed) max_speed
from
(
with w as
(
select y.vid, y.date_time, y.speed, y.status, y.over_30, y.next_time, decode(y.next_time_over_30, y.next_time, 'N', 'Y') end_of_block
from
(
select x.vid, x.date_time, x.speed, x.status, x.over_30, x.next_time, lead(x.date_time, 1, null) over (partition by x.vid order by x.date_time) next_time_over_30
from
(
select vs.vid, vs.date_time, vs.speed, vs.status, case when vs.speed >= 30 then 'Y' else 'N' end over_30, lead(vs.date_time, 1, null) over (partition by vs.vid order by vs.date_time) next_time
from veh_speed vs
) x
where x.over_30 = 'Y'
) y
)
select w1.vid, w1.date_time, w1.speed, w1.status, w1.over_30, w1.next_time, w1.end_of_block, min(w2.date_time) end_time
from w w1, w w2
where w2.end_of_block = 'Y'
and w2.date_time >= w1.date_time
group by w1.vid, w1.date_time, w1.speed, w1.status, w1.over_30, w1.next_time, w1.end_of_block
order by w1.vid, w1.date_time
) z
group by z.vid, z.end_time
;
这给出了：
VID START_TIME END_TIME MAX_SPEED
1 Jan-15-2014 10:00:35 Jan-15-2014 10:00:50 38
1 Jan-15-2014 10:00:15 Jan-15-2014 10:00:25 45