MySQL: Select rows, grouped by ID, where none of the rows have a value in a certain column

Question

Confusing title, I know. Ii couldn't think of a better way to word it. Basically, I have a table like this:

OwnerID | LotID | CurrentOwner

The OwnerID is the unique key for the table. This value is never repeated.
The LotID IS repeated. This table stores historical owner records for lots.
The CurrentOwner column simply contains a Y or NULL for each row. It is possible for one LotID to have multiple owners.

What I need is basically just for reporting. I need to know which LotIDs DO NOT have a "Y" for CurrentOwner in any of their rows. If I had the following data:

1, 1, Y  
2, 1, NULL  
3, 1, NULL  
4, 2, NULL  
5, 2, NULL  
6, 3, Y  
7, 3, NULL

Then I would need to know that LotID #2 does not have any owners currently.

Answer 1

I was able to solve my problem with the following query:

SELECT SUM(CASE 
             WHEN o.CurrentOwner IS NULL THEN 1
             ELSE 0
           END) AS numNull,
       COUNT(*) AS total,
       LotID
         FROM owner o GROUP BY LotID HAVING numNull=total

Answer 2

Redesign the schema. Have a separate table with PRIMARY KEY(lotID) that has a column OwnerID. This eliminates the chance of having two owners for one lot.

It does not, however, eliminate the issue of a lot with no owner. But the solution is pretty simple with either LEFT JOIN or NOT EXISTS.

Meanwhile, with the current schema, this should be a concise solution:

SELECT LotID
    FROM owner
    GROUP BY LotID
    HAVING  COALESCE(CurrentOwner) IS NULL;