Question

我有一张类似于

的表格

create table LOCHIST
(
  RES_ID VARCHAR(10) NOT NULL,
  LOC_DATE TIMESTAMP NOT NULL,
  LOC_ZONE VARCHAR(10)
)

使用

等值

insert into LOCHIST values(0911,2015-09-23 12:27:00.000000,SYLVSYLGA);
insert into LOCHIST values(5468,2013-02-15 13:13:24.000000,30726);
insert into LOCHIST values(23894,2013-02-15 13:12:13.000000,BECTFOUNC);
insert into LOCHIST values(24119,2013-02-15 13:12:09.000000,30363);
insert into LOCHIST values(7101,2013-02-15 13:11:37.000000,37711);
insert into LOCHIST values(26083,2013-02-15 13:11:36.000000,SHAWANDAL);
insert into LOCHIST values(24978,2013-02-15 13:11:36.000000,38132);
insert into LOCHIST values(26696,2013-02-15 13:11:27.000000,29583);
insert into LOCHIST values(5468,2013-02-15 13:11:00.000000,37760);
insert into LOCHIST values(5552,2013-02-15 13:10:55.000000,30090);
insert into LOCHIST values(24932,2013-02-15 13:10:48.000000,JBTTLITGA);
insert into LOCHIST values(23894,2013-02-15 13:10:42.000000,47263);
insert into LOCHIST values(26803,2013-02-15 13:10:25.000000,32534);
insert into LOCHIST values(24434,2013-02-15 13:10:03.000000,PLANSUFVA);
insert into LOCHIST values(26696,2013-02-15 13:10:00.000000,GEORALBGA);
insert into LOCHIST values(5468,2013-02-15 13:09:54.000000,19507);
insert into LOCHIST values(23894,2013-02-15 13:09:48.000000,37725);

这个表确实持续了数百万条记录。

每个RES_ID表示将其位置ping到LOC_ZONE的预告片的ID，然后在LOC_DATE中存储。

我想要找到的是特定位置区域中所有预告片花费的平均时间。例如，如果拖车x在loc区域PLANSUFVA中花费了4个小时，并且拖车y在loc区域PLANSUFVA中花了6个小时我想要返回

Loc Zone  Avg Time  
PLANSUFVA   5

无论如何都没有游标吗？

我非常感谢你的帮助。

Answer 1

这需要SQL 2012：

with data
as (
      select *, (case when LOC_ZONE != PREVIOUS_LOC_ZONE or PREVIOUS_LOC_ZONE is null then ROW_ID else null end) as STAY_START, (case when LOC_ZONE != NEXT_LOC_ZONE or NEXT_LOC_ZONE is null then ROW_ID else null end) as STAY_END
      from (
            select RES_ID, LOC_ZONE, LOC_DATE, lead(LOC_DATE, 1) over (partition by RES_ID, LOC_ZONE order by LOC_DATE) as NEXT_LOC_DATE, lag(LOC_ZONE, 1) over (partition by RES_ID order by LOC_DATE) as PREVIOUS_LOC_ZONE, lead(LOC_ZONE, 1) over (partition by RES_ID order by LOC_DATE) as NEXT_LOC_ZONE, ROW_NUMBER() over (order by RES_ID, LOC_ZONE, LOC_DATE) as ROW_ID
            from LOCHIST
      ) t
), stays  as (
      select * from (
            select RES_ID, LOC_ZONE, STAY_START, lead(STAY_END, 1) over (order by ROWID) as STAY_END
            from (
                  select RES_ID, LOC_ZONE, STAY_START, STAY_END, ROW_NUMBER() over (order by RES_ID, LOC_ZONE, STAY_START desc) as ROWID
                  from data
                  where STAY_START is not null or STAY_END is not null 
            ) t
      ) t
      where STAY_START is not null and STAY_END is not null
)
select s.LOC_ZONE, avg(datediff(second, LOC_DATE, NEXT_LOC_DATE)) / 60 / 60 as AVG_IN_HOURS
from data d
inner join stays s on d.RES_ID = s.RES_ID and d.LOC_ZONE = s.LOC_ZONE and d.ROW_ID >= s.STAY_START and d.ROW_ID < s.STAY_END
group by s.LOC_ZONE

Answer 2

要解决此问题，您需要花费在每个位置的时间。

执行此操作的一种方法是使用相关子查询。您需要对相邻值进行分组。我们的想法是找到序列中的下一个值：

select resid, min(loc_zone) as loc_zone, min(loc_date) as StartTime,
       max(loc_date) as EndTime,
       nextdate as NextStartTime
from (select lh.*,
             (select min(loc_date) from lochist lh2
              where lh2.res_id = lh.res_id and lh2.loc_zone <> lh.loc_zone and
                    lh2.loc_date > lh.loc_date
             ) as nextdate
      from lochist lh
     ) lh
 group by lh.res_id, nextdate

使用此数据，您可以获得所需的平均值。

我不清楚时间是否应该基于EndTime - StartTime（最后记录的时间减去第一个记录的时间）或NextStartTime - startTime（第一次在下一个位置减去第一次在这个位置）。

此外，这会为每个res_id的最后一个位置返回NULL。你没有说明如何处理序列中的最后一个。

如果您在res_id, loc_date, loc_zone上构建索引，它可能会运行得更快。

如果您使用的是Oracle或SQL Server 2012，则正确的查询是：

select lh.*,
       lead(loc_date) over (partition by res_id order by loc_date) as nextdate
from (select lh.*,
             lag(loc_zone) over (partition by res_id order by loc_date) as prevzone
      from lochist lh
     ) lh
where prevzone is null or prevzone <> loc_zone

现在每次入住有一行，而nextdate是下一个区域的日期。

Answer 3

要在不使用游标或相关子查询的情况下执行此操作，请尝试：

with rl as
(select l.*, rank() over (partition by res_id order by loc_date) rn
 from lochist l),
fdr as
(select rc.*, coalesce(rn.loc_date, getdate()) next_date
 from rl rc
 left join rl rn on rc.res_id = rn.res_id and rc.rn + 1 = rn.rn)
select loc_zone, avg(datediff(second, loc_date, next_date))/3600 avg_time
from fdr
group by loc_zone

SQLFiddle here.

（由于SQLServer计算时差的方式，计算平均时间（以秒为单位）然后除以60 * 60可能更好。除了getdate（）和datediff子句 - 可以用{替换{1}}和sysdate - 这应该在SQLServer 2005及以后的Oracle 10g中都有效。）

Answer 4

这应该按照平均花费的分钟数来获得每个区域。 CROSS APPLY返回不同区域中的下一个ping。

SELECT
     loc.LOC_ZONE
    ,AVG(DATEDIFF(mi,loc.LOC_DATE,nextPing.LOC_DATE)) AS avgMinutes
FROM LOCHIST loc
CROSS APPLY(
    SELECT TOP 1 loc2.LOC_DATE
    FROM LOCHIST loc2
    WHERE loc2.RES_ID = loc.RES_ID
    AND loc2.LOC_DATE > loc.LOC_DATE
    AND loc2.LOC_ZONE <> loc.LOC_ZONE
    ORDER BY loc2.LOC_DATE ASC
) AS nextPing
GROUP BY loc.LOC_ZONE
ORDER BY avgMinutes DESC

Answer 5

我的解决方案的变体：

select LOC_ZONE, avg(TOTAL_TIME) AVG_TIME from (
    select RES_ID, LOC_ZONE, sum(TIME_SPENT) TOTAL_TIME
    from (
        select RES_ID, LOC_ZONE, datediff(mi, lag(LOC_DATE, 1) over (
            partition by RES_ID order by LOC_DATE), LOC_DATE) TIME_SPENT
        from LOCHIST
    ) t
    where TIME_SPENT is not null
    group by RES_ID, LOC_ZONE) f
group by LOC_ZONE

这说明在同一地点多次停留。 lag或lead之间的选择取决于停留应该以ping开始还是结束（即，如果一个预告片从A发送ping，然后x小时后从B发送，那么这是A还是B）。

根据位置区域的变化计算平均时间花费

5 个答案: