sql - 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 EmailApproved = 1 should not be selected.

declare @Test table (id int, EmailApproved bit, Created datetime);

insert into @Test (id, EmailApproved, Created)
values
  (1,0,'2011-03-07 03:58:58.423')
  , (2,0,'2011-02-21 04:55:52.103')
  , (3,0,'2011-01-29 13:24:02.103')
  , (4,1,'2010-10-12 14:41:54.217')
  , (5,0,'2010-10-12 14:34:15.903')
  , (6,0,'2010-10-12 10:10:19.123')
  , (7,1,'2010-08-27 12:07:16.073')
  , (8,1,'2010-08-25 12:15:49.413')
  , (9,0,'2010-08-25 12:14:51.970')
  , (10,1,'2010-04-12 16:43:44.777');

select *
  , case when Row1 = Row2 then 1 else 0 end Row1EqualRow2
from (
  select id, EmailApproved, Created
    , row_number() over (partition by EmailApproved order by Created desc) Row1
    , row_number() over (order by Created desc) Row2
  from @Test
) X
--where Row1 = Row2
order by Created desc;

Which produces the following results:

id  EmailApproved   Created                 Row1    Row2    Row1EqualsRow2
1   0               2011-03-07 03:58:58.423 1       1       1
2   0               2011-02-21 04:55:52.103 2       2       1
3   0               2011-01-29 13:24:02.103 3       3       1
4   1               2010-10-12 14:41:54.217 1       4       0
5   0               2010-10-12 14:34:15.903 4       5       0
6   0               2010-10-12 10:10:19.123 5       6       0
7   1               2010-08-27 12:07:16.073 2       7       0
8   1               2010-08-25 12:15:49.413 3       8       0
9   0               2010-08-25 12:14:51.970 6       9       0
10  1               2010-04-12 16:43:44.777 4       10      0

What I actually want is:

id  EmailApproved   Created                 Row1    Row2    Row1EqualsRow2
1   0               2011-03-07 03:58:58.423 1       1       1
2   0               2011-02-21 04:55:52.103 2       2       1
3   0               2011-01-29 13:24:02.103 3       3       1
4   1               2010-10-12 14:41:54.217 1       4       0
5   0               2010-10-12 14:34:15.903 4       5       0
6   0               2010-10-12 10:10:19.123 5       6       0

Note: Row, Row2 & Row1EqualsRow2 are just working columns to show my calculations.

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;