substring after split by a separator oracle - sql

Question

Like if I have a string "123456,852369,7852159,1596357"
The out put looking for "1234,8523,7852,1596"
Requirement is....we want to collect 4 char after every ',' separator
like split, substring and again concat
select
REGEXP_REPLACE('MEDA,MEDA,MEDA,MEDA,MEDA,MEDA,MEDA,MEDA,MDCB,MDCB,MDCB,MDCB,MDCB,MDCB', '([^,]+)(,\1)+', '\1')
from dual;

Answer 1

we want to collect 4 char after every ',' separator
Here is an approach using regexp_replace:
select regexp_replace(
'123456,852369,7852159,1596357',
'([^,]{4})[^,]*(,|$)',
'\1\2'
)
from dual
Regexp breakdown:
([^,]{4}) 4 characters others than "," (capture that group as \1)
[^,]* 0 to n characters other than "," (no capture)
(,|$) either character "," or the end of string (capture this as \2)
The function replaces each match with capture 1 (the 4 characters we want) followed by capture 2 (the separator, if there is one).
Demo:
RESULT
1234,8523,7852,1596

Answer 2

One option might be to split the string, extract 4 characters and aggregate them back:
SQL> with test (col) as
2 (select '123456,852369,7852159,1596357' from dual)
3 select listagg(regexp_substr(col, '[^,]{4}', 1, level), ',')
4 within group (order by level) result
5 from test
6 connect by level <= regexp_count(col, ',') + 1;
RESULT
--------------------------------------------------------------------------------
1234,8523,7852,1596
SQL>

Answer 3

With REGEX_REPLACE:
select regexp_replace(the_string, '(^|,)([^,]{4})[^,]*', '\1\2')
from mytable;
This looks for
the beginning of the string or the comma
then four characters that are not a comma
then any number of trailing characters that are not a comma
And only keeps
the beginning or the comma
the four characters that follow
Demo: https://dbfiddle.uk/efUFvKyO