Select top rows until value in specific column has appeared twice

Question

I have the following query where I am trying to select all records, ordered by date, until the second time EmailApproved = 1 is found. The second record where <

Answer 1

Steps:

1. Create a row number, rn, over all rows in case id is not in sequence.
2. Create a row number, approv_rn, partitioned by EmailApproved so we know when EmailApproved = 1 for the second time
3. Use a outer apply to find the row number of the second instance of EmailApproved = 1
4. In the where clause filter out all rows where the row number is >= the value found in step 3.
5. If there is 1 or 0 EmailApproved records available then the outer apply will return null, in which case return all available rows.

    ; with test as
    (
        select  *, 
                rn         = row_number() over (order by Created desc),
                approv_rn  = row_number() over (partition by EmailApproved 
                                                    order by Created desc)
        from    @Test
    )
    select  *
    from    test t
            outer apply
            (
                select  x.rn
                from    test x
                where   x.EmailApproved = 1
                and     x.approv_rn     = 2
            ) x
    where   t.rn    < x.rn or x.rn is null
    order by t.Created desc