Selecting entries with biggest value less than list of values - sql

Question

Suppose my table structure as follows:
id | Word
---|-----
1 | a
2 | aa
. | ..
I have a list of id's like this:
(...,900, 1000, 2000, 3000, 4000,....)
I want to find the biggest id less than each id in the above list.
My table id's is not necessarily consecutive and there are some gaps between two successive id's, for example:
(...,889,900,950,952,997,1000,1001,1010,1920,2000,2990,3000,3500,4000,...)
The expected result according to the above list would be:
(889, 997, 1920, 2990, 3500,...)
How do i achieve desired results?

Answer 1

Use a common table expression and ROW_NUMBER()
;WITH cte AS(
SELECT *, ROW_NUMBER() OVER (ORDER BY ID) rowNum
FROM example)
SELECT ID, word
FROM cte
WHERE rowNum IN (
SELECT (rowNum - 1)
FROM cte
WHERE ID IN ('900','1000','2000','3000','4000'))
--WHERE ID IN (SELECT ID FROM <tableWithIDs>))
If you already have all of the ID you are looking for in another table, you would instead use the commented portion of my answer instead of the hardcoded IN list.
This will work only if the ID you are looking for exists in the table. So, as noted in a comment below if you were searching for 1001 you would not get 997, unless 1001 existed in the table (meaning, if it existed it would get a rowNum value and could be used to decrement in the subquery)
[DEMO HERE]
The following is another way to just see what the previous ID is for each row:
SELECT *, LEAD(ID,1) OVER(ORDER BY ID DESC) PreviousID
FROM example
ORDER BY ID

Answer 2

I would simply do:
select v.val, t.*
from (values (900), (1000), (2000), (3000), (4000) ) v(val) outer apply
(select top 1 t.*
from t
where t.id < v.val
order by t.id desc
) t;
This allows you to see the value on each of the rows. That is probably important because SQL result sets are unordered and it will not be obvious which value goes with which row.
EDIT:
If you know the row numbers are in the table, the most performance solution is probably:
select t.*
from (select t.*, lead(id) over (order by id) as next_id
from t
) t
where next_id in ( . . . );

Answer 3

This should work and I think it will be fairly efficient.
declare #V table (num int primary key);
insert into #V values (800), (889), (900), (997), (1000), (1910), (1920), (2000), (2990), (3000), (3500), (4000);
declare #T table (num int primary key);
insert into #T values (800), (900), (1000), (1200), (2000), (3000), (4000);
select tt.vP
from ( select t.num as t, v.num as v
, LAG(v.num) over (order by v.num) as vP
from #V v
left join #T t
on v.num = t.num
) tt
where tt.t is not null
and tt.vP is not null
order by tt.vP
Not clear how you want it to behave
select t.num
, (select max(v.num) from #V v where v.num < t.num) as prior
from #T t