Oracle PL-SQL “Regular expression” to replace each “(space)and(space)” with ',' in a string

Question

I have a string like this x=y and g=h and u=1 and I need to replace each (space)and(space) with ,. I'm using the powerful regular_expression for this but it does not give me the desired result.

select regexp_replace('x=y and g=h and u=1','(^[[:space:]]*)AND(^[[:space:]]*)', ',')
from dual;

I was wondering if you could help me out here. thanks in advance.

Answer 1

If the format is always exactly one space either side of a lowercase and, then you don't need regexp_replace as plain old replace will do it:

with demo (str) as
      ( select 'x=y and g=h and u=1' from dual )
select str
     , replace(str, ' and ', ',') as plain_replace
from   demo;

STR                 PLAIN_REPLACE
------------------- -------------------
x=y and g=h and u=1 x=y,g=h,u=1

A regex is needed if the number of spaces can vary:

with demo (str) as
      ( select 'x=y  and g=h and   u=1' from dual )
select str
     , replace(str, ' and ', ',') as plain_replace
     , regexp_replace(str, '\s+and\s+', ',') as one_or_more_spaces
from   demo;

STR                     PLAIN_REPLACE          ONE_OR_MORE_SPACES
----------------------- ---------------------- ---------------------------
x=y  and g=h and   u=1  x=y ,g=h,  u=1         x=y,g=h,u=1

The + means one or more, in contrast to * which means zero or more. For example I assume you don't want Band in Spandex replaced with B,in Sp,ex.

To be on the safe side, I've used the Perl-style \s (whitespace) operator rather than a literal space character, as this will also match any tab characters.

Answer 2

select regexp_replace('x=y and g=h and u=1','([[:space:]]*)and([[:space:]]*)', ',')
from dual;

REGEXP_REPL
-----------
x=y,g=h,u=1