Attain value present in earlier set but missing in later set - sql

Question

I can't get my head around why this is so difficult to achieve!
All I want to do is retrieve a value from an earlier set of results that does not exist in a later set of results from within the same table.
This is what I'm trying to achieve:
select b.Id
from table a
where
not exists
(select b.Id from table b where b.Id = a.Id and
b.time = (select max(time) from table where time < a.time))
I know this is not syntactically correct, the difficulty is accessing the b.Id while maintaining a reference to the other set's time and it's an issue in every form I've tried, such as using except. I've also tried outer joins to the same table to try to find the missing Id but to no avail.
I found it easy to find the Id of a missing row present in the later set and not the previous, but not the other way around.
I'm using SQL Server 2008R2.
EDIT
This is hard to pen so I'll give an example:
Table
- PK - Id - Time
------------------
- 1 - 100 - 13:00
- 2 - 99 - 13:00
- 3 - 100 - 11:00
- 4 - 99 - 11:00
- 5 - 98 - 11:00
- 6 - 100 - 10:00
- 7 - 99 - 10:00
- 8 - 98 - 10:00
- 9 - 97 - 10:00
So the result I expect would be rows:
{ 5, 98, 13:00 } and { 9, 97, 11:00 } as thse IDs are missing from the set that comes before. Note the time is from the later set, to show that they're missing from that set of time.

Answer 1

I would consider doing this with group by and having. If I understand correctly:
select t.id
from table t
group by t.id
having max(t.time) < #LaterTime and
sum(case when t.time >= #EarlierTime then 1 else 0 end) > 0;

Answer 2

Try to use table variables.
DECLARE #table1 TABLE (column1, column2, column3)
DECLARE #table2 TABLE (column1, column2, column3)
SELECT INTO #table1 (column1, column2, column3)
VALUES
FROM IntendedTable
WHERE Date < DesiredDate
SELECT INTO #table1 (column1, column2, column3)
VALUES
FROM IntendedTable
WHERE Date > DesiredDate
SELECT *
FROM #table1
WHERE NOT EXISTS (SELECT * FROM #table2)
The above isn't syntactically correct, but it would likely be the approach I would take.

Answer 3

Ids having exactly one time, the first one
select *
from table
where id in (
select t1.id
from table t1
group by t1.id
having count(distinct t1.time) = 1 )
and time = (select min(t2.time) from table t2)
EDIT
Following OP comments, all missing rows above min time for the id
select id,time
from (
select t1.id,t2.time
from
(select distinct id from table ) t1
cross join
(select distinct time from table) t2
except
select distinct id,time
from table
) tt
where time > (select min(t3.time) from table t3 where t3.id=tt.id)

Answer 4

I think this is what you want. Gives expected result.
CREATE TABLE #T (PK INT , ID INT , [TIME] TIME)
INSERT INTO #T
SELECT 1 ,100 , '13:00' UNION ALL
SELECT 2 ,99 , '13:00' UNION ALL
SELECT 3 ,100 , '11:00' UNION ALL
SELECT 4 ,99 , '11:00' UNION ALL
SELECT 5 ,98 , '11:00' UNION ALL
SELECT 6 ,100 , '10:00' UNION ALL
SELECT 7 ,99 , '10:00' UNION ALL
SELECT 8 ,98 , '10:00' UNION ALL
SELECT 9 ,97 , '10:00'
;WITH R
AS
(
SELECT DENSE_RANK() OVER (ORDER BY T1.TIME DESC) RANK,*
FROM #T T1
)
SELECT DISTINCT R1.PK, R1.ID, R2.TIME
FROM R R1
INNER JOIN R R2 ON R1.RANK = R2.RANK + 1
WHERE
NOT EXISTS
(
SELECT ID, TIME FROM R R2 WHERE R1.RANK = R2.RANK + 1 AND R1.ID = R2.ID
)