Oracle 11g - SQL за изчисляване на времевата разлика между няколко реда

Бележки.... не трябва да имате излишни данни (име, фамилия) в тази таблица, трябва да имате отделна таблица само за това. Изглежда, че часовете ви са съкратени, а не закръглени? (закръгляването ще даде 1,26 в първия ред).

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