Duplicates in the table

Question

hi,
how to find the duplicates in the table in the below example
Id     Empfirstname     Empfirstlastname     empdesig
1     Xyz     Abc     Software Engg
2     Xyz     Abc     Software Engg
3     Kkk     Ddd     Architect
need query to display
1     Xyz     Abc     Software Engg
2     Xyz     Abc     Software Engg
duplicate records

Answer 1

In addition:
You might want to think about the 'upper-/lowercase-thing' and still being/being not duplicate in these cases.
If so, then spot the difference:
MHO%xe> select * from
  2  (
  3  with all_your_data_are_belong_to_us -- generate some data on the fly here
  4  as (
  5      select 1 Id,  'Xyz' empfirstname,  'Abc' emplastname, 'Software Eng' empdesig from dual uni
on all
  6      select 2, 'Xyz', 'Abc', 'Software Eng' from dual union all
  7      select 3, 'aaa', 'AAA', 'Fairy' from dual union all
  8      select 4, 'AAA', 'aaa', 'Fairy' from dual union all
  9      select 5, 'Zlad', 'Molvania', 'Electrician' from dual union all
10      select 6, 'Kkk', 'Ddd', 'Architect' from dual
11     )
12  select id
13  ,      empfirstname
14  ,      emplastname
15  ,      empdesig
16  ,      count(*) over ( partition by upper(empfirstname), upper(emplastname), upper(empdesig)
17                         order by 'geronimo' ) rn
18  from   all_your_data_are_belong_to_us  --<< this could be YOUR table name, if so, omit the with-clause! ;-)
19  )
20  where rn > 1;
        ID EMPF EMPLASTN EMPDESIG             RN
         3 aaa  AAA      Fairy                 2
         4 AAA  aaa      Fairy                 2
         2 Xyz  Abc      Software Eng          2
         1 Xyz  Abc      Software Eng          2
Verstreken: 00:00:00.68
MHO%xe> select * from
  2  (
  3  with all_your_data_are_belong_to_us -- generate some data on the fly here
  4  as (
  5      select 1 Id,  'Xyz' empfirstname,  'Abc' emplastname, 'Software Eng' empdesig from dual uni
on all
  6      select 2, 'Xyz', 'Abc', 'Software Eng' from dual union all
  7      select 3, 'aaa', 'AAA', 'Fairy' from dual union all
  8      select 4, 'AAA', 'aaa', 'Fairy' from dual union all
  9      select 5, 'Zlad', 'Molvania', 'Electrician' from dual union all
10      select 6, 'Kkk', 'Ddd', 'Architect' from dual
11     )
12  select id
13  ,      empfirstname
14  ,      emplastname
15  ,      empdesig
16  ,      count(*) over ( partition by empfirstname, emplastname, empdesig
17                         order by 'geronimo' ) rn
18  from   all_your_data_are_belong_to_us  --<< this could be YOUR table name, if so, omit the with-clause! ;-)
19  )
20  where rn > 1;
        ID EMPF EMPLASTN EMPDESIG             RN
         1 Xyz  Abc      Software Eng          2
         2 Xyz  Abc      Software Eng          2( As you can see, I used Karthick's example, and it worked after translating the 'French' part (partitoin vs. partition) ;-) )