Regular Expression to Match string pattern followed by N digits in snowflake

Question

I have dataset like below in table A and I want result-set where the pattern for value column is like ABC followed by any 8 digits.

Output should be like below after regex match

[

I tried to use regex_sbustr but failed to get actual result set for 8 digits pattern.

Will appreciate your help

Answer 1

And to show @JNevil's REGEXP_LIKE solution working:

select 
    column1, 
    column2, 
    REGEXP_LIKE(column2, 'ABC[0-9]{8}.*') as match
from values
    (1, 'ABC123456788999'),
    (2, 'ABC123458765uhfh=hh'),
    (3, 'BCA123456788999'),
    (4, 'ABC987654321'),
    (5, 'ABC876hjkl90'),
    (6, 'ABC876');

gives:

COLUMN1	COLUMN2	MATCH
1	ABC123456788999	TRUE
2	ABC123458765uhfh=hh	TRUE
3	BCA123456788999	FALSE
4	ABC987654321	TRUE
5	ABC876hjkl90	FALSE
6	ABC876	FALSE

thus in filtering form:

select 
    column1, 
    column2 as value
from values
    (1, 'ABC123456788999'),
    (2, 'ABC123458765uhfh=hh'),
    (3, 'BCA123456788999'),
    (4, 'ABC987654321'),
    (5, 'ABC876hjkl90'),
    (6, 'ABC876')
where REGEXP_LIKE(column2, 'ABC[0-9]{8}.*'); 

gives:

COLUMN1	VALUE
1	ABC123456788999
2	ABC123458765uhfh=hh
4	ABC987654321

As per the documents, the LIKE version is automatically anchored, which means there is an implicit ^ and $ added to the begin/end of you regexp string, thus the need for the .* in this solution, otherwise none of the given input will match, as they all have 9 or more tokens.

as seen here (with extra 8 numeric input):

select 
    column1, 
    column2, 
    REGEXP_LIKE(column2, 'ABC[0-9]{8}') as match
from values
    (1, 'ABC123456788999'),
    (2, 'ABC123458765uhfh=hh'),
    (3, 'BCA123456788999'),
    (4, 'ABC987654321'),
    (5, 'ABC876hjkl90'),
    (6, 'ABC876'),
    (8, 'ABC12345678')
    ;    

COLUMN1	COLUMN2	MATCH
1	ABC123456788999	FALSE
2	ABC123458765uhfh=hh	FALSE
3	BCA123456788999	FALSE
4	ABC987654321	FALSE
5	ABC876hjkl90	FALSE
6	ABC876	FALSE
8	ABC12345678	TRUE