How do you return a grouped expression from REGEXP_SUBSTR?

Question

I have the string USERSbthompsonDESKTOP, and I want to return 'bthompson'.
Something equivalent to this Perl code:
my ($user) = $string =~ /USERS(.+?)DESKTOP/;
print "$user\n";
Output is 'bthompson'.
-- This does not work...
select regex_substr('USERSbthompsonDESKTOP', 'USERS(.+?)DESKTOP') from dual;
returns: USERSbthompsonDESKTOP
I tried adding a backreference, but don't know how it works.
select regexp_substr('USERSbthompsonDESKTOP', 'USERS(.+?)DESKTOP', '\1') from dual; -- Return the first expression?
returns:
ORA-01722: invalid number
01722. 00000 - "invalid number"
*Cause:   
*Action:
select regexp_substr('USERSbthompsonDESKTOP', 'USERS(.+?)DESKTOP\1') from dual;
returns Null.
Is there a way to return only 'bthompson' from that string?

Answer 1

Hi,
In Oracle 10, you can use REGEXP_REPLACE for that:
SELECT  REGEXP_REPLACE ( 'USERSbthompsonDESKTOP'
                 , '.*USERS(.+?)DESKTOP.*'
                 , '\1'
                 )          AS middle_part
FROM    dual;In Oracle 11, you can also use REGEXP_SUBSTR:
SELECT  REGEXP_SUBSTR ( 'USERSbthompsonDESKTOP'
                , 'USERS(.+?)DESKTOP'
                , 1
                , 1
                , NULL
                , 1
                )          AS middle_part
FROM    dual;