Extract string between different special symbols - sql

Question

I am having following string in my query
.\ABC\ABC\2021\02\24\ABC__123_123_123_ABC123.txt
beginning with a period from which I need to extract the segment between the final \ and the file extension period, meaning following expected result
ABC__123_123_123_ABC123
Am fairly new to using REGEXP and couldn't help myself to an elegant (or workable) solution with what Q&A here or else. In all queries the pattern is the same in quantity and order but for my growth of knowledge I'd prefer to not just count and cut.

Answer 1

You can use REGEXP_REPLACE function such as
REGEXP_REPLACE(col,'(.*\\)(.*)\.(.*)','\2')
in order to extract the piece starting from the last slash upto the dot. Preceding slashes in \\ and \. are used as escape characters to distinguish the special characters and our intended \ and . characters.
Demo

Answer 2

You need just regexp_substr and simple regexp ([^\]+)\.[^.]*$
select
regexp_substr(
'.\ABC\ABC\2021\02\24\ABC__123_123_123_ABC123.txt',
'([^\]+)\.[^.]*$',
1, -- position
1, -- occurence
null, -- match_parameter
1 -- subexpr
) substring
from dual;
([^\]+)\.[^.]*$ means:
([^\]+) - find one or more(+) any characters except slash([] - set, ^ - negative, ie except) and name it as group \1(subexpression #1)
\. - then simple dot (. is a special character which means any character, so we need to "escape" it using \ which is an escape character)
[^.]* - zero or more any characters except .
$ - end of line
So this regexp means: find a substring which consist from: one or more any characters except slash followed by dot followed by zero or more any characters except dot and it should be in the end of string. And subexpr parameter = 1, says oracle to return first subexpression (ie first matched group in (...))
Other parameters you can find in the doc.

Answer 3

Here is my simple full compatible example with Oracle 11g R2, PCRE2 and some other languages.
Oracle 11g R2 using function substr (Reference documentation)
select
regexp_substr(
'.\ABC\ABC\2021\02\24\ABC__123_123_123_ABC123.txt',
'((\w)+(_){2}(((\d){3}(_)){3}){1}((\w)+(\d)+){1}){1}',
1,
1
) substring
from dual;
Pattern: ((\w)+(_){2}(((\d){3}(_)){3}){1}((\w)+(\d)+){1}){1}
Result: ABC__123_123_123_ABC123
Just as simple as it can be, regular expressions always follow a minimal standard, as you can see portability also provided, just for the case someone else is interested in going the simplest way.
Hopefully, this will help you out!