Regular Expression - Quick Question

Question

Hi Experts,
SELECT REGEXP_SUBSTR('123@123','[^@@]+',1,1) FROM DUAL;
Result:
123 -- Wrong one.
SELECT REGEXP_SUBSTR('123@@123','[^@@]+',1,1) FROM DUAL;
Result:
123 -- Right one.How can i achieve this?
Thanks,

Answer 1

michaels2 wrote:
SQL> with t as
select '123@123' str from dual union
select '123@@123'  from dual
select str, regexp_replace(str,'@@.*') x  from t
STR      X      
123@123  123@123
123@@123 123    
2 rows selected.
No michael, he needs something like:
SQL> with t as
  2  (
  3   select '123@@456@@789@@000' str from dual
  4  )
  5  select regexp_substr(replace(str,'@@','#'),'[^#]+',1,level)
  6  from t
  7  connect by level<=(length(str)-length(replace(str,'@@','')))/2+1;
REGEXP_SUBSTR(R
123
456
789
000Because he has to extract tokens separated by @@, but I can't find a neat solution...
seems like Oracle is ignoring the grouping operator:
SQL> with t as
  2  (select '123@123' str from dual union
  3  select '123@@123'  from dual)
  4  select str, regexp_substr(str,'[^(@@)]+')
  5  from t;
STR      REGEXP_S
123@123  123
123@@123 123In the first row it should extract the full string, not just the first three characters...
Max
[My Italian Oracle blog| http://oracleitalia.wordpress.com/2010/02/07/aggiornare-una-tabella-con-listruzione-merge/]