Opmerkingen.... je zou geen overbodige gegevens (voornaam, achternaam) in deze tabel moeten hebben, daar zou je een aparte tabel voor moeten hebben. Het lijkt erop dat uw uren worden afgekapt en niet afgerond? (de afronding zou 1,26 opleveren in de eerste rij).
with
test_data ( seq, the_date, scanid, locatn, user_id, first_name, last_name ) as (
select 103939758, to_date('05-NOV-16 14:36:22', 'dd-MON-yy hh24:mi:ss'), 194972, 'DOOR 19', 'AX9868', 'Mike', 'Derry' from dual union all
select 103939780, to_date('05-NOV-16 14:38:07', 'dd-MON-yy hh24:mi:ss'), 194972, 'DOOR 19', 'AX9868', 'Mike', 'Derry' from dual union all
select 103939792, to_date('05-NOV-16 14:39:24', 'dd-MON-yy hh24:mi:ss'), 194972, 'DOOR 19', 'AX9868', 'Mike', 'Derry' from dual union all
select 103940184, to_date('05-NOV-16 15:16:53', 'dd-MON-yy hh24:mi:ss'), 194972, 'DOOR 19', 'AX9868', 'Mike', 'Derry' from dual union all
select 103940185, to_date('05-NOV-16 15:51:41', 'dd-MON-yy hh24:mi:ss'), 194972, 'DOOR 19', 'AX9868', 'Mike', 'Derry' from dual union all
select 103940214, to_date('05-NOV-16 09:51:42', 'dd-MON-yy hh24:mi:ss'), 194993, 'DOOR 16', 'BC1910', 'Tony', 'McCann' from dual union all
select 103940215, to_date('05-NOV-16 15:19:06', 'dd-MON-yy hh24:mi:ss'), 194993, 'DOOR 16', 'BC1910', 'Tony', 'McCann' from dual
)
-- end of test data; solution (SQL query) begins below this line
select trunc(the_date) as the_date, user_id, first_name, last_name,
trunc(24 * (max(the_date) - min(the_date)), 2) as total_hrs
from test_data
group by trunc(the_date), user_id, first_name, last_name
;
THE_DATE USER_ID FIRST_NAME LAST_NAME TOTAL_HRS
--------- ------- ---------- --------- ----------
05-NOV-16 AX9868 Mike Derry 1.25
05-NOV-16 BC1910 Tony McCann 5.45