How to extract group from regular expression in Oracle?

I got this query and want to extract the value between the brackets.

select de_desc, regexp_substr(de_desc, '\[(.+)\]', 1)
from DATABASE
where col_name like '[%]';

It however gives me the value with the brackets such as "[TEST]". I just want "TEST". How do I modify the query to get it?

The third parameter of the REGEXP_SUBSTR function indicates the position in the target string (de_desc in your example) where you want to start searching. Assuming a match is found in the given portion of the string, it doesn't affect what is returned.

In Oracle 11g, there is a sixth parameter to the function, that I think is what you are trying to use, which indicates the capture group that you want returned. An example of proper use would be:

SELECT regexp_substr('abc[def]ghi', '\[(.+)\]', 1,1,NULL,1) from dual;

Where the last parameter 1 indicate the number of the capture group you want returned. Here is a link to the documentation that describes the parameter.

10g does not appear to have this option, but in your case you can achieve the same result with:

select substr( match, 2, length(match)-2 ) from (
SELECT regexp_substr('abc[def]ghi', '\[(.+)\]') match FROM dual
);

since you know that a match will have exactly one excess character at the beginning and end. (Alternatively, you could use RTRIM and LTRIM to remove brackets from both ends of the result.)

You need to do a replace and use a regex pattern that matches the whole string.

select regexp_replace(de_desc, '.*\[(.+)\].*', '\1') from DATABASE;