using regex_substr to get corresponding value from another column in oracle - oracle

Question

I have the below query which works, which gives the nth string corresponding to the key I give in the where clause as names (separator being ##)
select names from (
select
regexp_substr('a##b##c##d##e##f','[^##]+', 1, level) as names,
rownum as nth
from dual
connect by regexp_substr('a##b##c##d##e##f', '[^##]+', 1, level) is not null
)
where nth in (
select nth from (
select
regexp_substr('150##200##13##8##51##61','[^##]+', 1, level) as names,
rownum as nth
from dual
connect by regexp_substr('150##200##13##8##51##61', '[^##]+', 1, level) is not null
)
where names = '200'
)
Now, I have a table temp with say 3 columns x,y and z where x has strings like a##b##c##d##e##f, y has 1##2##3##4##5##6 and z will have number like 1.
If I have a rows like
a##b##c##d##e##f 150##200##13##8##51##61 200
a##b##c##d##e##f 1##2##3##4##5##6 2
g##h##i##j##k##l 1##2##3##4##5##99 99
I want outputs like
a##b##c##d##e##f 150##200##13##8##51##61 200 b
a##b##c##d##e##f 1##2##3##4##5##6 2 b
g##h##i##j##k##l 1##2##3##4##5##99 99 l
simply plugging "temp" in place of dual in the above query takes long time as the db has over 50k rows. Any better solution or how do I fix this?

Answer 1

You may find a plain SQL solution, but I'd prefer to capsule the critical substring functionality in a function. After that the query is trivial
update
with tcols as (
select rownum colnum from dual connect by level <= 6 /* (max) number of columns */),
t2 as (
select x,y,z, colnum,
nth_substring(y,'#',colnum) subs
from regexp_tst, tcols
)
select
x,y,z, colnum,
nth_substring(x,'#',colnum) a
from t2
where subs = z
;
.
X Y Z A
---------------- ---------------- ---------- ----
a##b##c##d##e##f 1##2##3##4##5##6 1 a
a##b##c##d##e##f 1##2##3##4##5##6 2 b
g##h##i##j##k##l 1##2##3##4##5##6 3 i
The required function is as follows (You may want to adjust the trimming and the repeated delimieter logic)
create or replace function nth_substring( i_str VARCHAR2, i_del VARCHAR2, i_pos NUMBER)
/*
find n-th substring in delimited string
i_str input string e.g. a##b##c##d##e##f
i_del delimiter character
i_pos position of the strin starting from 1
*/
return VARCHAR2 is
BEGIN
return rtrim(ltrim(regexp_substr(i_str,'[^'||i_del||']+', 1, i_pos)));
END;
/