Oracle REGEXP_SUBSTR - SEMICOLON STRING EXTRACTION - regex

Question

I have below input and need mentioned output. How I can get it. I tried different pattern but could not get through it.
so in brief, any value having all three 1#2#3 parts(if it is present) or first value should be returned
2#9#;2#37#65 -> 2#37#65
2#9#;2#37#65;2#37# -> 2#37#65
2#9#;2#37#65;2#37#;2#37#56 -> 2#37#65 or 2#37#56
2#37#65;2#99 -> 2#37#65
3#9#;3#37#65;3#37#36;2#37#56 -> 3#37#65 or 3#37#36 or 2#37#56
2#37#;2#99# -> 2#37 or 2#99# ( in this case any value)
I tried few patterns and other pattern but no help.
regexp_substr('2#9#;2#37#65;2#37#','#[^;]+',1)
SUBSTR(REGEXP_SUBSTR(SUBSTR(uo_filiere,1,INSTR(uo_filiere,';',1)-1), '#[^#]+$'),2)

Answer 1

You can use a REGEXP_REPLACE here:
REGEXP_REPLACE(uo_filiere, '^(.*;)?([0-9]+(#[0-9]+){2,}).*|^([^;]+).*', '\2\4')
See the regexp demo
Details:
^ - start of string
(.*;)? - an optional Group 1 capturing any text and then a ;
([0-9]+(#[0-9]+){2,}) - Group 2 (\2): one or more digits, and then two or more occurrences of # followed with one or more digits
.* - the rest of the string
| - or
^([^;]+).* - start of string, Group 4 capturing one or more chars other than ; and then any text till end of string.
The replacement is Group 2 + Group 4 values.

Answer 2

Here is a simple-minded way to solve this. It may prove more efficient than other approaches, given the particular nature of the problem.
First, use a regular expression to find the first token that has all three parts. This part of the solution should be the most efficient approach for those strings that do have a three-part token, and it performs work that must be performed on all input strings in any case.
In the second part, wrap within nvl - if no three-part token is found, select the first token regardless of how many parts are present. This part uses only substr and instr in a trivial manner, so it should be very fast too.
Here's the query, run on a few more sample inputs to test those cases too.
with
sample_data (uo_filiere) as (
select '2#9#;2#37#65' from dual union all
select '2#9#;2#37#65;2#37#' from dual union all
select '2#9#;2#37#65;2#37#;2#37#56' from dual union all
select '2#37#65;2#99' from dual union all
select '3#9#;3#37#65;3#37#36;2#37#56' from dual union all
select '2#37#;2#99#' from dual union all
select '1#22#333' from dual union all
select '33#444#' from dual
)
select uo_filiere,
nvl(regexp_substr(uo_filiere, '(;|^)(([^;#]+#){2}[^;]+)', 1, 1, null, 2)
, substr(uo_filiere, 1, instr(uo_filiere || ';', ';') - 1)
) as first_value
from sample_data
;
UO_FILIERE FIRST_VALUE
---------------------------- ----------------------------
2#9#;2#37#65 2#37#65
2#9#;2#37#65;2#37# 2#37#65
2#9#;2#37#65;2#37#;2#37#56 2#37#65
2#37#65;2#99 2#37#65
3#9#;3#37#65;3#37#36;2#37#56 3#37#65
2#37#;2#99# 2#37#
1#22#333 1#22#333
33#444# 33#444#