Группировка по двум кейсам с датой

Question

Я должен сгруппировать по дате для запроса ниже, но часы REG и OT идут в разных строках на одну и ту же дату. Может кто-нибудь сказать мне, как этого добиться?

Я сделал группу по дате, но она не работает.

SELECT VP.PERSONNUM AS PERSONID, 
VP.FIRSTNM AS FIRST_NAME, 
VP.LASTNM AS LAST_NAME, 
(select personnum from vp_employeev42 where personnum = vp.homelaborlevelnm6) as FOREMAN_CODE, 
CA.PERSONCSTMDATATXT as SHIFT, 
TO_CHAR(X.APPLYDTM, 'DD-MM-YYYY') AS LABOR_DATE, 
VP.HOMELABORLEVELNM3 AS DEPARTMENT, 
CASE WHEN TS.LABORLEVELNAME3 != 'NULL' THEN TS.LABORLEVELNAME3 ELSE VP.HOMELABORLEVELNM3 END AS CHGD_DEPARTMENT, 
CASE WHEN P.NAME = 'US-Regular' THEN SUM(X.DURATIONSECSQTY/3600.0) ELSE 0 END AS REG, 
CASE WHEN P.NAME = 'US-Overtime' THEN SUM(X.DURATIONSECSQTY/3600.0) ELSE 0 END AS OT, 
CASE WHEN P.NAME = 'US-Doubletime' THEN SUM(X.DURATIONSECSQTY/3600.0) ELSE 0 END AS DBL , 
CASE WHEN TS.LABORLEVELNAME4 != 'NULL' THEN TS.LABORLEVELNAME4 ELSE VP.HOMELABORLEVELNM4 END AS ACCOUNT_ID 
FROM VP_EMPLOYEEV42 VP, 
WFCTOTAL X, PAYCODE P, 
PERSONCSTMDATA CA, 
VP_TIMESHTPUNCHV42 TS 
WHERE VP.PERSONID = X.EMPLOYEEID AND 
X.PAYCODEID = P.PAYCODEID AND 
X.EMPLOYEEID = CA.PERSONID AND 
CA.CUSTOMDATADEFID ='13' AND 
X.TIMESHEETITEMID = TS.TIMESHEETITEMID AND 
P.NAME in ('US-Regular','US-Overtime','US-Doubletime') AND 
X.EMPLOYEEID in (select personid from PERSONCSTMDATA where CUSTOMDATADEFID ='14' and PERSONCSTMDATATXT = 'USKEANE') 
and VP.PAYRULENAME in ('USKEANE 10 HR Shift','USKEANE 12 HR Shift','USKEANE 1st Shift','USKEANE 2nd Shift','USKEANE 3rd Shift') 
and X.APPLYDTM between '01-APR-19' and '30-APR-19' 
group by VP.PERSONID, VP.PERSONNUM, VP.FIRSTNM, VP.LASTNM, VP.HOMELABORLEVELNM6, 
X.APPLYDTM, VP.HOMELABORLEVELNM3, CA.PERSONCSTMDATATXT ,VP.HOMELABORLEVELNM4, 
P.NAME, TS.LABORLEVELNAME3, TS.LABORLEVELNAME4 ORDER BY VP.PERSONID, X.APPLYDTM DESC

{
PERSONID FIRST_NAME LAST_NAME FOREMAN_CODE SHIFT LABOR_DATE DEPARTMENT CHGD_DEPARTMENT REG OT DBL ACCOUNT_ID
E0002887 Maria Putnam E0061775 1 10-04-2019 25855 25855 0 4 0 60467
E0002887 Maria Putnam E0061775 1 10-04-2019 25855 25855 4 0 0 60467
E0002887 Maria Putnam E0061775 1 15-04-2019 25855 25855 8 0 0 60467
}

Answer 1

Вы делаете это:

select case p.name when 'US-Regular'    then sum(x.durationsecsqty/3600) else 0 end as reg, 
       case p.name when 'US-Overtime'   then sum(x.durationsecsqty/3600) else 0 end as ot, 
       case p.name when 'US-Doubletime' then sum(x.durationsecsqty/3600) else 0 end as dbl
  from p, x
  where x.paycodeid = p.paycodeid 
  group by p.name

Но вы не хотите различать p.name в вашем выводе.Поэтому удалите его из group by и измените case часть, как здесь:

select sum(case p.name when 'US-Regular'    then x.durationsecsqty/3600 else 0 end) as reg, 
       sum(case p.name when 'US-Overtime'   then x.durationsecsqty/3600 else 0 end) as ot, 
       sum(case p.name when 'US-Doubletime' then x.durationsecsqty/3600 else 0 end) as dbl
  from p, x
  where x.paycodeid = p.paycodeid 

На вашем месте я бы также изменил синтаксис старого соединения на более читаемый:

  ...
  from paycode        p  
  join wfctotal       x  on x.paycodeid = p.paycodeid 
  join personcstmdata ca on x.employeeid = ca.personid 
  ...
  where p.name in ('US-Regular','US-Overtime','US-Doubletime') 
  ...

хотя есть люди, которые предпочитают старый способ:)