Help in finding a regex pattern

Question

Dear all REGEX Gurus. 
We have a comma separated string like this:
132, 143, "222, 144, abc", 227, 888, "222#55#ab"
As you might have guessed, this comes from excel when we save as CSV file, and when one of the columns in excel has the value 222,144,abc. So excel itself puts " at beginning and end.
Now we want to split it based on ',' like this:
132
143
222,144,abc
227
888
222#55#ab
So what we thought is to use a REGEX to find a pattern that has anything beginning with ", ending with ", and has a comma (,) in between. We'd replace that comma with a special character.
System seems to perform greedy search if i search like this ",".
I tried this:
"([^"]+)" -> this works that it gives me all values within quotes. But I want only those which have a comma in them.
So in above, i do not want 222#55#ab to come.
Tried various combinations, but not able to get it work.
Can someone advise please how to achieve it?
Thanks in adv.

Answer 1

Hi,
I am in 4.6 version and dont have the facility to regex it for you.
Just wrote a sample code, if useful then use it.
DATA:lv_string(100) TYPE c.
DATA:lv_len TYPE i.
TYPES:BEGIN OF ty,
      field(100) TYPE c,
      END OF ty.
DATA:it TYPE TABLE OF ty.
FIELD-SYMBOLS:<fs> TYPE ty.
lv_string = '132, 143, "222, 144, abc", 227, 888, "222#55#ab"'.
CONDENSE lv_string NO-GAPS.
SPLIT lv_string AT '"' INTO TABLE it.
CLEAR lv_string.
LOOP AT it ASSIGNING <fs>.
  IF <fs> IS INITIAL.
    delete it index sy-tabix.
    CONTINUE.
  ENDIF.
  lv_len = strlen( <fs> ) - 1.
  IF ( <fs>+lv_len(1) CA '"' ) OR ( <fs>+lv_len(1) CA ',' ).
    <fs>+lv_len(1) = ' '.
  ENDIF.
  IF ( <fs>+0(1) CA '"' ) OR ( <fs>+0(1) CA ',' ).
    <fs>+0(1) = ' '.
  ENDIF.
  CONDENSE <fs>.
  WRITE:/ <fs>.
ENDLOOP.