Use REGEXP_INSTR to find a text string with space(s) in it

Question

I am trying to use REGEXP_INSTR to find a text string with space(s) in it.
(This is in a Function.)
Let's say ParmIn_Look_For has a value of 'black dog'. I want to see if
ParmIn_Search_This_String has 'black dog' anywhere in it. But it gives an error
Syntax error on command line.
If ParmIn_Look_For is just 'black' or 'dog' it works fine.
Is there some way to put single quotes/double quotes around ParmIn_Look_For so this will
look for 'black dog' ??
Also: If I want to use the option of ignoring white space, is the last parm
'ix' 'i,x' or what ?
SELECT
REGEXP_INSTR(ParmIn_Search_This_String,
'('||ParmIn_Look_For||')+', 1, 1, 0, 'i')
INTO Position_Found_In_String
FROM DUAL;
Thanks, Wayne

Answer 1

Maybe something like this ?
test@ORA10G>
test@ORA10G> with t as (
  2    select 1 as num, 'this sentence has a black dog in it' as str from dual union all
  3    select 2, 'this sentence does not' from dual union all
  4    select 3, 'yet another dog that is black' from dual union all
  5    select 4, 'yet another black dog' from dual union all
  6    select 5, 'black dogs everywhere...' from dual union all
  7    select 6, 'black dog running after me...' from dual union all
  8    select 7, 'i saw a black dog' from dual)
  9  --
10  select num, str
11  from t
12  where regexp_like(str,'black dog');
       NUM STR
         1 this sentence has a black dog in it
         4 yet another black dog
         5 black dogs everywhere...
         6 black dog running after me...
         7 i saw a black dog
5 rows selected.
test@ORA10G>
test@ORA10G>pratz
Also, 'x' ignores whitespace characters. Link to doc:
http://download.oracle.com/docs/cd/B19306_01/server.102/b14200/conditions007.htm#i1048942
Message was edited by:
pratz