Question

每天查询自行车的首次旅行持续时间和最后一次旅行持续时间。

Table
trip_id bike-id trip_date   trip_starttime    trip_duration
1          1    2018-12-01  12:00:00.0000000    10
2          2    2018-12-01  14:00:00.0000000    25
3          1    2018-12-01  14:30:00.0000000    5
4          3    2018-12-02  05:00:00.0000000    12
5          3    2018-12-02  19:00:00.0000000    37
6          1    2018-12-02  20:30:00.0000000    20

Expected Result

trip_date   bike-id first_trip_duration last_trip_duration
2018-12-01    1            10               5
2018-12-01    2            25              25
2018-12-02    1            20              20
2018-12-02    3            12              37

我用以下代码尝试过

select A.trip_date,A.[bike-id],A.trip_duration AS Minduration,B.trip_duration AS MaxDUrtaion from 

(SELECT T1.trip_date,T1.[bike-id],T1.trip_duration FROM TRIP T1
INNER JOIN (
select trip_date,[bike-id] , min(trip_starttime) AS Mindate 
from Trip  group by trip_date,[bike-id] ) T2
oN T1.[bike-id]=T2.[bike-id] AND T1.trip_date=T2.trip_date AND t1.trip_starttime=t2.Mindate ) as A
inner join 

(SELECT T1.trip_date,T1.[bike-id],T1.trip_duration FROM TRIP T1
INNER JOIN (
select trip_date,[bike-id] , MAX(trip_starttime) AS Maxdate 
from Trip  group by trip_date,[bike-id] ) T2
oN T1.[bike-id]=T2.[bike-id] AND T1.trip_date=T2.trip_date AND t1.trip_starttime=t2.Maxdate )  as B

ON A.[bike-id]=B.[bike-id] AND A.trip_date=B.trip_date

order by A.trip_date,A.[bike-id]

我也想知道其他一些逻辑，请帮忙。

Answer 1

select trip_date,bike_id
      ,first_value(trip_duration) over(partition by trip_date,bike_id order by trip_starttime) as first_trip_duration
      ,first_value(trip_duration) over(partition by trip_date,bike_id order by trip_starttime desc) as last_trip_duration
from trip;

Answer 2

首先，为每个日期/自行车确定第一次和最后一次旅行。
然后，确定这些旅行的持续时间。

类似的事情可以做到（尽管我没有测试）：

SELECT minmax.trip_date,
       minmax.bike_id,
       first.trip_duration AS first_trip_duration,
       last.trip_duration AS last_trip_duration
FROM   (SELECT   trip_date,
                 bike_id,
                 MIN(trip_starttime) AS first_trip,
                 MAX(trip_starttime) AS last_trip
        FROM     trip_table
        GROUP BY trip_date,
                 bike_id
       ) minmax
JOIN   trip_table first
  ON   minmax.trip_date = first.trip_date
 AND   minmax.bike_id = first.bike_id
 AND   minmax.first_trip = first.trip_starttime
JOIN   trip_table last
  ON   minmax.trip_date = last.trip_date
 AND   minmax.bike_id = last.bike_id
 AND   minmax.last_trip = last.trip_starttime

假设表上有必要的索引。
最好是（bike_id，trip_date，starttime）的唯一索引。

Answer 3

假设支持窗口功能，可以通过first_value完成。

select distinct
       trip_date
      ,bike_id
      ,first_value(trip_duration) over(partition by trip_date,bike_id order by trip_starttime) as first_trip_duration
      ,first_value(trip_duration) over(partition by trip_date,bike_id order by trip_starttime desc) as last_trip_duration
from trip

每天找出自行车的第一次旅行持续时间和最后一次旅行持续时间

3 个答案: