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. )
18. AND
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.