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Oracle 11g - SQL per calcolare la differenza di tempo tra più righe

Note.... non dovresti avere dati ridondanti (nome, cognome) in questa tabella, dovresti avere una tabella separata solo per quello. Sembra che le tue ore siano troncate e non arrotondate? (l'arrotondamento darebbe 1,26 nella prima riga).

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