Counting types of characters using REGEXP

Question

Hi,
Is there a simle way to count the types of characters per words in a sentence using regexp?
I'm trying to get the number of upper and lowercase chars for every [:lower:] and[:upper:] patterns:
For example
'John Bell' -> 'A(1)a(3) A(1)a(3)'
'Frederic McLaughing' -> 'A(1)a(7) A(1)a(1)A(1)a(7)'
Thanks for your help
-S.

Answer 1

MODEL clause to the rescue! You read my mind Volder. ;-) Though I didn't want to work on a MODEL solution until after I knew for sure the OP could use it. Too many times they end up saying "but I'm on Oracle 8" after the fact.
In hindsight I should have guessed they had 10g because of the fact they asked for a REGEXP solution.
Regardless, here's what I was thinking, though it doesn't seem as efficient as your approach.
set null "(null)"
column s format a30
column t format a30
with t as
  ( select 'John Bell' str from dual union all
    select 'Frederic McLaughing' from dual union all
    select null from dual
select str, s from t
model
partition by (str)
dimension by (0 dim)
measures
( regexp_replace(regexp_replace(str,'[[:lower:]]','x'),'[[:upper:]]','X') s )
rules iterate(100) -- set this to maximum word size
( s[0] =
     replace
     ( s[0],
       rpad( 'X', 100-iteration_number, 'X' ),
       'A(' || to_char(100-iteration_number) || ')'
   s[0] =
     replace
     ( s[0],
       rpad( 'x', 100-iteration_number, 'x' ),
       'a(' || to_char(100-iteration_number) || ')'
order by 1
STR                 S
Frederic McLaughing A(1)a(7) A(1)a(1)A(1)a(7)
John Bell           A(1)a(3) A(1)a(3)
(null)              (null)--
Joe Fuda
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