Use Regex from a column in Redshift - regex

Question

I have 2 tables in Redshift, one of them has a column containing Regex strings. And I want to join them like so:
select *
from one o
join two t
on o.value ~ t.regex
But this query throws an error:
[Amazon](500310) Invalid operation: The pattern must be a valid UTF-8 literal character expression
Details:
-----------------------------------------------
error: The pattern must be a valid UTF-8 literal character expression
code: 8001
context:
query: 412993
location: cgx_impl.cpp:1911
process: padbmaster [pid=5211]
-----------------------------------------------;
As far as I understood from searching in the docs, the right side of a regex operator ~ must be a string literal.
So this would work:
select *
from one o
where o.value ~ 'regex'
And this would fail:
select *
from one o
where 'regex' ~ o.value
Is there any way around this? Anything I missed?
Thanks!

Answer 1

Here's a workaround I am using. Maybe it's not super fast, but it works:
First create a function:
CREATE FUNCTION is_regex_match(pattern text, s text) RETURNS BOOLEAN IMMUTABLE AS $$
import re
return True if re.search(pattern, s) else False
$$ LANGUAGE plpythonu;
Then use it like this (o.value contains a regex pattern):
select *
from one o
where is_regex_match(o.value, 'some string');

Answer 2

You could try using the built-in function regexp_substr()
https://docs.aws.amazon.com/redshift/latest/dg/REGEXP_SUBSTR.html
select *
from one o
join two t
on regexp_substr(o.value, t.regex) <> ''
Edit example added of raw query
It appears that the fields must be explicitly cast as varchars when built.
with fake_table as (
SELECT 'sample value'::varchar as value, '[a-z]'::varchar as pattern
)
SELECT *
, regexp_substr(value, pattern)
FROM
fake_table
WHERE
regexp_substr(value, pattern) <>''