REGEXP_REPLACE: How to use a function over the found strings?

Question

Hello,
Consider the following:
select regexp_replace('A1BBCCA2BBCC', '(A.)', '[\1]')
from dual
'[A1]BBCC[A2]BBCC'Now I try to put a function on the replaced strings:
select regexp_replace('A1BBCCA2BBCC', '(A.)', lower('[\1]'))
from dual
'[A1]BBCC[A2]BBCC'The result is the same i.e. the function lower has been executed with my reg. expression as a parameter, not the result strings. How can I execute the function passing as a parameter not the reg. expression but the strings, that have been found? (Of course, my real need is to use a custom function, not "lower").
Thanks in advance.
Best Regards,
Beroetz

Answer 1

I'm sure there must be a simpler way, but this is my first thought... (although strictly speaking in this example I could get rid of the lower function and just include 'a' hardcoded.)
SQL> ed
Wrote file afiedt.buf
  1  with t as (select 'A1BBCCA2BBCC' as txt from dual)
  2  -- end of data
  3  select replace(sys_connect_by_path(lower('A'||substr(x,1,1))||substr(x,2),','),',') as x
  4  from (
  5        select REGEXP_SUBSTR(txt, '[^A.]+', 1, level) as x, level rn
  6        from t
  7        connect by level <= length(regexp_replace(txt,'[^A.]*'))
  8       )
  9  where connect_by_isleaf = 1
10  connect by rn = prior rn+1
11* start with rn = 1
SQL> /
X
a1BBCCa2BBCC
SQL>