Oracle - Regular expression - Keep reducing a char to match to another column - regex

Question

I have 2 columns from 2 different tables - say columnA and columnB, which I am matching with each other. However, if they do not match then I want to remove last one char from columnB and again match with columnA. If it still won't match then reduce one more char at the end from columnB and try to match. Keep reducing chars from columnB till there is match ( and untill columnB turns to 0 length).
Ex - ColumnA has a value "ABC" and columnB has "ABCDEF".
Then, since "ABC" is not equal to "ABCDEF", try to match "ABCDE" with "ABC". Since it is not matching then try "ABCD" . Since there is still no match then try "ABC" . Now there is match and so stop !!
I am unable to come with a regular expression in Oracle to handle this. I can use substr/length and bunch of "OR" conditions but I will prefer to avoid that if there is regular expression, which can do it nicely.
Thanks in advance.

Answer 1

SELECT *
FROM table_name
WHERE REGEXP_LIKE( columnb, '^'||columna||'.*$' );
(However, this has issues when columna contains ^$.*+?|[]{}()\ characters).
or
SELECT *
FROM table_name
WHERE columnb LIKE columna||'%';
or
SELECT *
FROM table_name
WHERE INSTR( columnb, columna ) = 1;
or
SELECT *
FROM table_name
WHERE SUBSTR( columnb, 1, LENGTH( columna ) ) = columna;

Answer 2

My guess is may be you want to find the longest prefix of two strings.
In my opinion, it's easier to do in PL/SQL than in SQL:
create or replace function longest_prefix(a varchar2, b varchar2) return varchar2 as
l number :=least(length(a), length(b));
l_common varchar2(32767) :=substr(a,1,l);
begin
for i in 1..l loop
if substr(a,i,1)!=substr(b,i,1) then
l_common:=substr(a,1,i-1);
exit;
end if;
end loop;
return l_common;
end;
/
Test:
SQL> select longest_prefix('asdf', 'as23') from dual;
LONGEST_PREFIX('ASDF','AS23')
--------------------------------------------------------------------------------
as