REGEXP_LIKE Oracle equivalent to count characters in Snowflake - oracle

Question

I am trying to come up with an equivalent of the below Oracle statement in Snowflake. This would check if the different parts of the string separated by '.' matches the number of characters in the REGEXP_LIKE expression. I have come up with a rudimentary version to perform the check in Snowflake but I am sure there's a better and cleaner way to do it. I am looking to come up with a one-liner regular expression check in Snowflake similar to Oracle. Appreciate your help!
-- Oracle
SELECT -- would return True
CASE
WHEN REGEXP_LIKE('AB.XYX.12.34.5670.89', '^\w{2}\.\w{3}\.\w{2}') THEN 'True'
ELSE NULL
END AS abc
FROM DUAL
-- Snowflake
SELECT -- would return True
REGEXP_LIKE(SPLIT_PART('AB.XYX.12.34.5670.89', '.', 1), '[A-Z0-9]{2}') AND
REGEXP_LIKE(SPLIT_PART('AB.XYX.12.34.5670.89', '.', 2), '[A-Z0-9]{3}') AND
REGEXP_LIKE(SPLIT_PART('AB.XYX.12.34.5670.89', '.', 3), '[A-Z0-9]{2}') AS abc

Answer 1

You need to add a .* at the end as the REGEXP_LIKE adds explicit ^ && $ to string:
The function implicitly anchors a pattern at both ends (i.e. '' automatically becomes '^$', and 'ABC' automatically becomes '^ABC$'). To match any string starting with ABC, the pattern would be 'ABC.*'.
select
column1 as str,
REGEXP_LIKE(str, '\\w{2}\\.\\w{3}\\.\\w{2}.*') as oracle_way
FROM VALUES
('AB.XYX.12.34.5670.89')
;
gives:
STR
ORACLE_WAY
AB.XYX.12.34.5670.89
TRUE
Or in the context of your question:
SELECT IFF(REGEXP_LIKE('AB.XYX.12.34.5670.89', '\\w{2}\\.\\w{3}\\.\\w{2}.*'), 'True', null) AS abc;

Answer 2

Your use of \w seems to suggest you don't need delimited strings to be strictly [A-Z0-9] since word characters allow underscore and period. If all bets were off and the only requirement was to have . at 3rd, 7th and 10th position, you could have used like this way.
select 'AB.XGH.12.34.5670.89' like '__.___.__.%' ;