Oracle SQL : Regexp_substr

Question

I have below sample values in a column

Abc-123-xyz
Def-456-uvw
Ghi-879-rst-123
Jkl-abc

Expected output is the third element split by '-', in case there is no third element, the last element will be retrieve.

See expected output below:

Xyz
Uvw
Rst
Abc

Thanks ahead for the help.

Answer 1

SELECT initcap(nvl(regexp_substr(word, '[^-]+', 1,3),regexp_substr(word, '[^-]+', 1,2)))  FROM your_table;

Answer 2

Another approach:

SQL> with t1(col) as(
  2    select 'Abc-123-xyz'     from dual union all
  3    select 'Def-456-uvw'     from dual union all
  4    select 'Ghi-879-rst-123' from dual union all
  5    select 'Jkl-Abc'         from dual
  6  )
  7  select regexp_substr( col
  8                      , '[^-]+'
  9                      , 1
 10                      , case
 11                           when regexp_count(col, '[^-]+') >= 3
 12                           then 3
 13                           else regexp_count(col, '[^-]+')
 14                        end
 15                      ) as res
 16    from t1
 17  ;

Result:

RES
---------------
xyz
uvw
rst
Abc

Answer 3

regexp_substr(column, '(.*?-){0,2}([^-]+)', 1, 1, '', 2)

Answer 4

You can also do it without RegEx:

with t1 as(
  select 'Abc-123-xyz' as MyText     from dual union all
  select 'Def-456-uvw'     from dual union all
  select 'Ghi-879-rst-123' from dual union all
  select 'Jkl-Abc'         from dual
)
SELECT 
  SUBSTR(t1.mytext, LENGTH(t1.mytext) - INSTR(REVERSE(t1.mytext), '-') + 2) 
FROM t1
;