How to select bottom most rows?

SELECT
    columns
FROM
(
     SELECT TOP 200
          columns
     FROM
          My_Table
     ORDER BY
          a_column DESC
) SQ
ORDER BY
     a_column ASC

Logically,

BOTTOM (x) is all the records except TOP (n - x), where n is the count; x <= n

E.g. Select Bottom 1000 from Employee:

In T-SQL,

DECLARE 
@bottom int,
@count int

SET @bottom = 1000 
SET @count = (select COUNT(*) from Employee)

select * from Employee emp where emp.EmployeeID not in 
(
SELECT TOP (@count-@bottom) Employee.EmployeeID FROM Employee
)

Sorry, but I don't think I see any correct answers in my opinion.

The TOP x function shows the records in undefined order. From that definition follows that a BOTTOM function can not be defined.

Independent of any index or sort order. When you do an ORDER BY y DESC you get the rows with the highest y value first. If this is an autogenerated ID, it should show the records last added to the table, as suggested in the other answers. However:

• This only works if there is an autogenerated id column
• It has a significant performance impact if you compare that with the TOP function

The correct answer should be that there is not, and cannot be, an equivalent to TOP for getting the bottom rows.

It is unnecessary. You can use an ORDER BY and just change the sort to DESC to get the same effect.