Removing LEADING and TRAILING keywords from a String in PL/SQL - regex

Question

I need to remove certain keywords from an input string and return the new string. Keywords are stored in another table like MR, MRS, DR, PVT, PRIVATE, CO, COMPANY, LTD, LIMITED etc. They are two kind of keywords LEADING - MR, MRS, DR and TRAILING - PVT, PRIVATE, CO, COMPANY, LTD, LIMITED etc.
So if Keywords is a LEADING then we have to remove that from the beginning and if it's a TRAILING then we have to remove that from the end. e.g.-MR Jones MRS COMPANY should return JONES MRS and MR MRS Jones PVT COMPANY should return MRS JONES PVT (As in first iteration MR and PVT will be trimmed and then word will become MRS JONES PVT) It should remove only very first occurrence of the reserve keyword either at the beginning or at the end of the input string so there are multiple occurrence of the LEADING keyword at the begining it should remove only the first one not the others like I gave example above, it is same for TRAILING keywords as well.
I have written the function below, and it is working fine but it is not efficient and I believe performance of this can be improved a lot(may be using regular expression). Below is the function:
CREATE OR REPLACE FUNCTION replace_keyword (p_in_name IN VARCHAR2)
RETURN VARCHAR2
IS
l_name VARCHAR2 (4000);
l_keyword_found BOOLEAN;
CURSOR c IS
SELECT *
FROM RSRV_KEY_WORDS
WHERE ACTIVE = 'Y'
AND upper(POSITION) in ('LEADING', 'TRAILING');
BEGIN
--Remove the leading and trailing blank spaces
l_name := TRIM (UPPER (p_in_name));
--remove LEADING keywords
l_keyword_found := false;
for rec in c LOOP
IF UPPER (rec.POSITION) = 'LEADING'
AND SUBSTR(l_name, 1,INSTR(l_name,' ',1) - 1) = rec.key_word
AND l_keyword_found = false
THEN
l_name := SUBSTR(l_name,INSTR(l_name,' ',1)+1);
l_keyword_found := true;
END IF;
EXIT WHEN (l_keyword_found);
END LOOP;
--Remove multiple spaces in a word and replace with single blank space
l_name := REGEXP_REPLACE (l_name, '[[:space:]]{2,}', ' ');
--Remove the leading and trailing blank spaces
l_name := TRIM (l_name);
--remove TRAILING keywords
l_keyword_found := false;
for rec in c LOOP
IF UPPER (rec.POSITION) = 'TRAILING'
AND SUBSTR(l_name, INSTR(l_name,' ',-1) + 1) = rec.key_word
AND l_keyword_found = false
THEN
l_name := SUBSTR(l_name,1,INSTR(l_name,' ',-1)-1);
l_keyword_found := true;
END IF;
EXIT WHEN (l_keyword_found);
END LOOP;
--Remove multiple spaces in a word and replace with single blank space
l_name := REGEXP_REPLACE (l_name, '[[:space:]]{2,}', ' ');
--Remove the leading and trailing blank spaces
l_name := TRIM (l_name);
return l_name;
EXCEPTION
WHEN OTHERS
THEN
raise_application_error (
-20001,
'An error was encountered - ' || SQLCODE || ' -ERROR- ' || SQLERRM);
END;
/

Answer 1

I cant really say if this will be faster, but I would give it a try:
Assuming the keywords in RSRV_KEY_WORDS does not change very often I would create a function to produce a regular expression from the table and have Oracle cache the result:
create or replace function get_lead_and_trail_regexp return varchar2
result_cache relies_on (RSRV_KEY_WORDS) is
declare
CURSOR c IS
SELECT ( SELECT listagg(key_word,'|') within group (order by 1)
FROM RSRV_KEY_WORDS
WHERE ACTIVE = 'Y'
AND upper(POSITION) = 'LEADING' ) as leading,
( SELECT listagg(key_word,'|') within group (order by 1)
FROM RSRV_KEY_WORDS
WHERE ACTIVE = 'Y'
AND upper(POSITION) = 'TRAILING' ) as trailing
FROM dual;
begin
for rec in c loop
return '(^[ ]+(('||rec.leading||')[ ]+))|([ ]+(('||rec.trailing||'||)[ ]+)$)';
end loop;
return null; -- Not very likely
end get_lead_and_trail_regexp;
You can then use the regular expression to remove first leading and first trailing keywords in one stroke:
l_name := REGEXP_REPLACE (l_name, get_lead_and_trail_regexp , ' ');
and then carry one with removing any duplicate spaces.
I have tested the regular expression with java.lang.String.replaceAll as I do not currently have an Oracle database available, but I believe it will work with REGEXP_REPLACE too.