SQL Regular expression greater than

Question

I have a table, lets say table A, with data following same pattern:

• ab1.c
• ab2.c
• ab3.c
• ab3.d

How could I Select data when the number is greater than 1, for example?

Right know, I am using:

SELECT * FROM A WHERE col LIKE "%a.b.%"

And then, in java, I manage to get which ones are greater than ab1

Isn't possible to do this in a single query?

(I need to do it in MySql and SQLServer)

Answer 1

If I understand your question correctly, you want to skip all rows where ab1.* and select data starting from ab2.*, so ab2 will be the first possible match for you, and query should look like this:

WITH dataset
AS
(
    SELECT 'a.b.1.c' AS List
    UNION ALL
    SELECT 'a.b.2.c'
    UNION ALL
    SELECT 'a.b.3.c'
    UNION ALL
    SELECT 'a.b.3.d'
)
SELECT * FROM dataset
WHERE List >= 'a.b.2'