case statment

I am using below query but still getting wrong results.

Select keyid,CASE
WHEN keyid like '^[A-Za-z]*%' THEN 'invalid'
else 'valid'
end as test

from table

I am expecting below

Appreciate your help in advance.

Thanks

Gold ⭐️

I used CASE WHEN keyid RLIKE '^[A-Za-z]' THEN 'invalid' else 'valid' end as test. This is working !

PraskaviAuthorAnswer

Gold ⭐️

I used CASE WHEN keyid RLIKE '^[A-Za-z]' THEN 'invalid' else 'valid' end as test. This is working !

Translate

Sign up