Oracle string replacement using lookup tables - oracle10g

Question

Can someone please help me with this issue?
I have my client address stored in a table (table CT) as a string which could have some abbreviated data. Example: 123 RD. I have another look up table (table LK) which contains possible street names for the abbreviations. RD = ROAD.
I would like to create a new table which has the complete address for this string.
Input:
table CT:
Add1 Add2 Add3
------------------------
123 RD APT 2 BLDG 1
test DR null null
main RD null BLDG2
table LK:
abbreviation completestreet
----------------------------------
RD road
APT apartment
BLDG building
DR drive
I would like to join these two tables to achieve the following:
123 ROAD APARTMENT 2 BUILDING 1
test DRIVE null null
main ROAD null BUILDING 2

Answer 1

You probably want to find a solution that uses regexp_replace together with the regexp word boundaries ^| and |$:
with ct as (
select '123 RD' add1, 'APT 2' add2, 'BLDG 1' add3 from dual union all
select '123 3RD street' add1, 'APT 2' add2, 'BLDG 1' add3 from dual union all
select 'test DR' add1, '' add2, '' add3 from dual union all
select 'main RD' add1, '' add2, 'BLDG2' add3 from dual
),
lk as (
select 'RD' abbreviation, 'road' completestreet from dual union all
select 'APT' abbreviation, 'apartment' completestreet from dual union all
select 'BLDG' abbreviation, 'building' completestreet from dual union all
select 'DR' abbreviation, 'drive' completestreet from dual
),
recursion (add1, add2, add3, l) as (
select add1, add2, add3, 1 l from ct union all
select regexp_replace(add1, '(^| )' ||abbreviation || '( |$)', '\1' || completestreet || '\2'),
regexp_replace(add2, '(^| )' ||abbreviation || '( |$)', '\1' || completestreet || '\2'),
regexp_replace(add3, '(^| )' ||abbreviation || '( |$)', '\1' || completestreet || '\2'),
l+1
from recursion join (
select
row_number() over (order by abbreviation) r,
abbreviation,
completestreet
from lk
) lk
on l=r
)
select substrb(add1, 1, 30),
substrb(add2, 1, 30),
substrb(add3, 1, 30)
from recursion
where l=(select count(*)+1 from lk);