Oracle SQL: Searching for $ in Regexp - regex

Question

I want to search my response field to find any instance where a dollar sign $ is not followed by a numerical value. 1 or 2 spaces before a numerical value is ok, but there shouldn't be any text values following $.
I have the following query below:
SELECT * FROM RESPONSES
WHERE (regexp_like(response, '(\$)(\s){1,2}[^0-9]'));
This should be able to identify responses that have "$ NA". Most responses will contain a combination of $ followed by numeric values and $ by text values.
I've tried a couple of variations of the above query without any success. Any thoughts?

Answer 1

You can use this:
SELECT *
FROM dual
WHERE
SIGN(REGEXP_INSTR (RESPONSE, '(\$)(\s){2}[^0-9]'))=0

Answer 2

Include the space character in your negated character set
Since a space character qualifies as a non-digit character, a second space can give a "false positive" for the data set you want to find.
SCOTT#db>WITH smple AS (
2 SELECT
3 '23 dkf $ 1' response
4 FROM
5 dual
6 UNION ALL
7 SELECT
8 '23 dkfg gjg $ 4'
9 FROM
10 dual
11 UNION ALL
12 SELECT
13 '$ NA'
14 FROM
15 dual
16 ) SELECT
17 s.*
18 FROM
19 smple s
20 WHERE
21 ( REGEXP_LIKE ( s.response,
22 '\$\s{1,2}[^ 0-9]+' ) );
RESPONSE
----------
$ NA

Answer 3

You may use :
select * from responses where regexp_like(a, '^\$\s')
to get values begin with a $ sign and followed with at least one space as consecutive character(s).
with t as
(
select '$ 524' as a from dual union all
select '$524' as a from dual union all
select '$ s67e' as a from dual union all
select '# 67e' as a from dual union all
select '$s67e' as a from dual union all
select '$#67e' as a from dual
)
select * from t where regexp_like(a, '^\$\s')
A
----
$ 524
$ s67e
Demo