查询的目的是获取预订房间状态为3或预留30至45分钟状态2或1无保留状态1.预订房间位于RESEENH表中,每个预订位于ORD_NOARCHIVE表中,该表具有开始时间和结束时间保留。因此,对于每个预订房间,该查询检查当前时间是否有预订,还检查会议室的父母和孩子。如果孩子被保留,那么父母就会被封锁。 获取前50条记录需要10秒钟。
with cte as
(
SELECT DISTINCT R.syscode,
R.behcode,
R.syscode AS FK_RESERVATIONUNIT, (
CASE
WHEN R.TYPE = 3 THEN '1'
WHEN R.TYPE = 1 THEN '2'
ELSE NULL
END ) AS LOCATION_TYPE,
R.sysobjalg AS FK_PROPERTY,
MP.syscode AS FK_MEASUREMENTPOINT,
MP.fk_plc_occupancy_state AS FK_PLC_OCCUPANCY_STATE,
F.syscode AS FK_FLOOR,
R.transitiontime,
r.type,
r.is_compoundreservationunit,
r.is_archived,
MP.fk_person,
os.transitionperiod
FROM reseenh R
--left outer join ordtrantbl RSS
--ON RSS.reservationunisyscode = R.syscode
left outer join objond F
ON F.syscode = R.fk_floor
left outer join pln_measurementpoint MP
ON MP.fk_reservationunit = R.syscode
AND MP.is_primary_measurement_point = 'T',
pln_ordersetting os
)
select cte.syscode,cte.behcode,cte.FK_RESERVATIONUNIT,
(CASE
WHEN O.begindatetime_user IS NULL THEN '1' --GREEN
WHEN O.begindatetime_user - (Nvl(cte.transitiontime, ( cte.transitionperiod ))/1440 ) > current_date THEN '2' -- ORANGE
WHEN O.begindatetime_user + (Nvl(cte.transitiontime, ( cte.transitionperiod )) /1440 ) > current_date THEN '3' -- RED
ELSE '3'
END ) AS LOCAVAILABILITY_STATUS_CODE,
cte.LOCATION_TYPE,
cte.FK_PROPERTY,
Coalesce(O.sysmelder, cte.fk_person) AS FK_PERSON,
O.syscode AS FK_ORDER,
O.ref_bostate_userdefined AS FK_ORDER_STATE_USER,
O.fk_bostate AS FK_ORDER_STATE_SYSTEM,
FK_MEASUREMENTPOINT,FK_PLC_OCCUPANCY_STATE,FK_FLOOR
from cte left outer join ord_noarchive O on O.syscode in
( SELECT MAX(ord.syscode) KEEP (DENSE_RANK FIRST ORDER BY ord.begindatetime_user) OVER (PARTITION BY ord.sysreseenh )
FROM ord_noarchive ORD
WHERE ( ( (
current_date >= ( ORD.begindatetime_user - ( Nvl(cte.transitiontime, ( cte.transitionperiod ))/1440) )
AND (
current_date - ( Nvl(cte.transitiontime, (cte.transitionperiod )) / 1440 ) ) <=ORD.enddatetime_user )
OR ( (
current_date + ( (
CASE
WHEN (
cte.TYPE = 1 ) THEN 30
ELSE 45
END ) / 1440 ) ) >= ( ORD.begindatetime_user - (Nvl(cte.transitiontime, ( cte.transitionperiod))/1440 ) )
AND (
current_date - ( Nvl(cte.transitiontime, ( cte.transitionperiod )) / 1440 ) ) < ORD.enddatetime_user ) )
AND ORD.sysreseenh IN
(
SELECT fk_reservationunit_parent
FROM pln_reservationunit_rut
WHERE fk_reservationunit_child IN
(
SELECT fk_reservationunit_child
FROM pln_reservationunit_rut
WHERE cte.is_compoundreservationunit = 'T'
AND fk_reservationunit_parent = cte.syscode)
UNION
SELECT cte.syscode
FROM dual
UNION
SELECT
CASE
WHEN cte.is_compoundreservationunit = 'T' THEN fk_reservationunit_child
ELSE fk_reservationunit_parent
END
FROM pln_reservationunit_rut
WHERE (
cte.is_compoundreservationunit = 'T'
AND fk_reservationunit_parent = cte.syscode )
OR (
cte.is_compoundreservationunit = 'F'
AND fk_reservationunit_child = cte.syscode ))
AND ORD.fk_bostate IN
(
SELECT syscode
FROM pln_bostate
WHERE pnname IN ( 'Requested',
'Made',
'AdministrativelyCompleted' )
AND fk_bodefinition = ref_bodefinition)
AND ORD.sysreseenh = O.sysreseenh
))
WHERE cte.is_archived = 'F'
AND cte.TYPE IN ( 1,
3 )
AND cte.fk_floor=495
答案 0 :(得分:0)
没时间分析一个非常复杂的查询的细节,但我的总体感觉是你试图一次做太多事情。分开任务。在一个查询中检查房间可用性,并在单独的查询中检查子/父事物。
此外,您可以分析查询的执行计划,并查看它对其造成的影响。我怀疑(再次没有时间真正尝试理解你的查询)是在某些时候,任务的混合正在转化为多对多的关系,你有一个中间结果,这是一些行之间的行的交叉产品查询中的表格。