I have this table with random values and I want to select only those that end with "_" or "%" in the flavor column using the option "ESCAPE".
I have tried this way but it also brings me those that have "_" in the middle of the value.
and if I try this way, I do not return any rows.
select * from bonyurt where sabor like '%\_' escape '\'
What am I doing wrong?
Try it like this:
select * from bonyurt where sabor like '%/_' escape '/'
I hope it serves you!
In order to obtain the result of a query that meets one or another condition, the OR operator must be used.
Applying the aforementioned to your question the query would be as follows:
select * from bonyurt where sabor like '%/_' escape '/' OR sabor like '%/%' escape '/';
Another alternative would be to consult each filter separately and then join both results with the operator UNION ALL
select * from bonyurt where sabor like '%/_' escape '/'
union all
select * from bonyurt where sabor like '%/%' escape '/';
I hope it serves you.