Question

employees
e_id first last
1 John Smith
2 Bob Smith
3 Alex Smith
4 John Doe
5 Ron Doe

clockpunch
e_id    time    for adjustment
1   0650    in  early
3   0710    in  late
4   0725    in  early
1   1100    lunch   ---
2   1150    in  late
2   1400    lunch   ---
4   1320    out ---

I need a SINGLE query that will list all employee names, along with a count of how many times that user has been early, descending order by the count of times early. How would this be done ?

Answer 1

    select a.first,a.last,sum(case b.adjustment when 'early' then 1 else 0 end) as ct 
from employees a, clockpunch b where a.e_id=b.e_id
    group by a.first, a.last