Question

I need make such request: choose list of cities which include letter i and have appropriate number of employees. Also I need take to attention that some city could have 0 employee and that for some employee field DEBT could be null.

Table depart

debt    city
-------------
43      odesa
23      kiev
79      lviv
78      lviv
12      rivne

Table empl

ide     fn      ln      debt
----------------------------
3421    jed     trt     43
354     jed     res     43
43      ged     hjkhg   79
73      ghghg   gfgf    79
456     jkl     gdfg    
532     kkhg    vjv     23
45      ki      vt      
243     ki      vt      78

I wrote this query:

select depart.CITY, count (*) as numb 
from depart 
   inner join empl on  empl.DEBT=depart.DEBT
where depart.CITY like '%i%'
group by depart.CITY;

But I do not know how take care about that some city could have 0 employee (for example this request does not show city rivne which have 0 employee) and that for some employee field DEBT could be null.

I use oracle with toad.

Expected results

city    numb
kiev    1
rivne   0
lviv    3

Answer 1

To take care of criterion move where before group by. To allow for non-existing employees or null DEBT, use left outer join.

select depart.CITY, count (empl.DEBT) as numb 
  from depart 
  left join empl 
    on empl.DEBT = depart.DEBT
 where depart.CITY like '%i%'
 group by depart.CITY;

To get all employees grouped by department you would reverse the outer join:

select depart.CITY, count (empl.ide) as numb 
  from empl  
  left join depart
    on empl.DEBT = depart.DEBT
 -- Note: condition is now part of a join. This is required
 --       as part of outer join because otherwise left hand
 --       table row would be filtered out.
   and depart.CITY like '%i%'
 group by depart.CITY;

(Presumably ide is primary key of employee).
But this would not return departments without employees, and CITY would be null where department’s city would not match %i% or empl.DEBT would be null from the start.

To solve the problem, one might extend first query with union all designed to retrieve employees without departments. But there is a question: do we want only departmentless employees, or we consider or employees not working in a city whose name contains i departmentless also. I’ve opted for second possibility.

select depart.CITY, count (empl.DEBT) as numb 
  from depart 
  left join empl 
    on empl.DEBT = depart.DEBT
 where depart.CITY like '%i%'
 group by depart.CITY;
select '(unknown or unmatched city)', count (*) as numb
  from empl
  left join depart
    on empl.DEBT = depart.DEBT
   and depart.CITY like '%i%'
 where depart.DEBT is null;

If you need distinction between employees without departments and those who work in a city not containing i, you might use case:

select case when depart.CITY like '%i%'
            then depart.CITY
            when depart.DEBT is null
            then '(No department)'
            else '(City does not match %i%)'
        end as CITY,
       count (*) as numb
  from empl
  left join depart
    on empl.DEBT = depart.DEBT
 group by 
       case when depart.CITY like '%i%'
            then depart.CITY
            when depart.DEBT is null
            then '(No department)'
            else '(City does not match %i%)'
        end

This query will count employees from matching cities, unmatching cities and those who have no department.