Prikazi cijelu temu 08.06.2020 15:47
zxz Van mreze
Administrator
Registrovan od:03.02.2009
Lokacija:Tuzla


Predmet:Re: SQL upiti
PreuzmiIzvorni kôd (MySQL):
  1. SELECT  rv.id_employees
  2. ,r.name
  3. ,rv.Date
  4. ,rv.id_department
  5. ,o.id_office
  6. ,CONCAT(k.id_cityDepartmentOffice,rv.id_department) as kljuc
  7. FROM tbl_schedule as rv
  8. INNER JOIN tbl_employees  as r ON rv.id_employees=r.id_employees
  9. INNER JOIN tbl_calendar as k   ON k.date=rv.date
  10. INNER JOIN tbl_citydepartmentoffice as o   ON o.Id_cityDeparmentOffice=k.id_cityDepartmentOffice
  11. AND rv.id_department=o.id_department
  12. WHERE  o.id_office
  13. IN (SELECT o.id_office
  14. FROM tbl_schedule as rv
  15. INNER JOIN tbl_citydepartmentoffice as o   ON rv.id_department=o.id_department
  16. WHERE rv.id_employees=1001
  17. )
  19. rv.id_employees<>1001
Da budem iskren nisam dovoljno ni procitao.
Gore sam mislio da su prve dvije tabele 1 na 1.
Ovdje je problem kljuca za cetvrtu tabelu i ako je on
IdCityDepartmentOffice i
IdDepartment
onda bi ovo rjesenje trebalo biti dobro.
Nisam testirao.
Podrška samo putem foruma, jer samo tako i ostali imaju koristi od toga.