REGEXP_SUBSTR function equivalent in Oracle 9i?

Question

Is there an equivalent function for regexp_substr in Oracle 9i.

I know regexp_substr came in Oracle 10g onwards.

Trying to figure out a way if I can use the same function logic in Oracle 9i.

I have data like

0/6/03/19
0/6/3/19
0/1/3/09

I want to pick out the values separately by delimiting the string with / .

I tried using substr and instr but it won't be generic as the length of the string between slashes can change.

Answer 1

Unfortunately you can only use instr and substr combination as :

with t as
(
  select '0/6/03/19' as str from dual union all
  select '0/6/3/19' from dual
)
select substr(str,1,instr(str,'/',1,1)-1) str1,
       substr(str,instr(str,'/',1,1)+1,instr(str,'/',1,2)-instr(str,'/',1,1)-1) str2,
       substr(str,instr(str,'/',1,2)+1,instr(str,'/',1,3)-instr(str,'/',1,2)-1) str3,
       substr(str,instr(str,'/',1,3)+1,length(str)-instr(str,'/',1,3)) str4
  from t;

STR1    STR2    STR3    STR4
----    ----    ----    ----
 0       6       03      19
 0       6        3      19

P.S. If your DB's version is 9.2 then with .. as structure may be used as above.

you can also get the results row-wise(in unpivot manner) as :

with t as
(
 select '/'||str||'/' as str, ID
   from
   (
    select 1 as ID, '0/6/03/19' as str from dual union all
    select 2,'0/6/3/19' from dual
   )
)
select
      distinct ID, level as piece_nr,
      substr(str,instr(str,'/',1,level)+1,instr(str,'/',1,level+1)-instr(str,'/',1,level)-1)
      as piece_value
  from ( select * from t )
connect by level <= length(str)-length(replace(str,'/',''))-1
  order by ID, level;

ID  PIECE_NR    PIECE_VALUE
--  --------  -----------
1        1           0
1        2           6
1        3           03
1        4           19
2        1           0
2        2           6
2        3           3
2        4           19