Without using a regular expressions in oracle 9i

Question

dear all;
I have the following test data below
  insert into p
      (id)
    values
      ('\G1\G2'); 
  insert into p
      (id)
    values
      ('\A1\');
   insert into p
      (id)
    values
      ('\B1\B2\B3');
       insert into p
      (id)
    values
      ('\J1\J2\J3\J4');and this is the output I desire though
ID
G1
null
B2
J3the output is gotten by looking at the second to the last entry and extracting it....
so for \G1\G2 the second to the last entry was G1 hence we have G1
for \A1\ there is no second to the last entry and that is why it is null
for \B1\B2\B3 the second to the last entry was B2

Answer 1

Hi,
user13328581 wrote:
... the output is gotten by looking at the second to the last entry and extracting it....
so for \G1\G2 the second to the last entry was G1 hence we have G1
for \A1\ there is no second to the last entry and that is why it is nullSo backslashes at the end of id don't matter. Whether id is
'\AI',
'\A1\' or
'\A1\\\\\\\\\\\\'
there's only entry in id, and so you want to return NULL. Is that right?
Here's one way to do that in Oracle 9:
WITH    got_pos          AS
     SELECT     id
     ,     INSTR ( RTRIM (id, '\')
                , -1
                , 2) + 1     AS pos_from
     ,     INSTR ( RTRIM (id, '\')
                , -1
                , 1)     AS pos_to
     FROM     p
SELECT       id
,       CASE
           WHEN  pos_from     > 1
           THEN  SUBSTR ( id
                            , pos_from
                  , pos_to - pos_from
       END     AS penult
FROM       got_pos
ORDER BY  id
;