Complex Query where record is max in set - sql

Question

I have some data that looks like this
CandidateCategory
candidateCategoryId
candidateId
categoryId
I want to return all records where a specific categoryId is the most recent entry, this max(candidateCategoryId)
So if a candidate has 5 categories I want to get that record for say category 23 but only if that is the most recent category added, ie candidateCategoryId is higher than all others for that category.
Using MS SQL 2012
Sample data in format
candidateCategoryId candidateId categoryId
100 1 10
101 1 11
102 1 50
103 1 23
104 1 40
no result, 23 isn't the max candidateCategoryId
candidateCategoryId candidateId categoryId
200 2 20
201 2 31
202 2 12
203 2 23
return result, 23 is the max candidateCategoryId for this candidate.

Answer 1

Try getting the max CandidateCategoryID per CandidateID First, then re-join back
select
yd2.*
from
( select yd.candidateID,
max( yd.candidateCategoryId ) as maxCandCatID
from YourData yd
group by yd.candidateID ) MaxPerID
JOIN YourData yd2
on MaxPerID.candidateID = yd2.candidateID
AND MaxPerID.maxCandCatID = yd2.CandidateCategoryID
AND yd2.categoryID = 23
So, from your sample data, the inner prequery "MaxPerID" will generate two rows...
CandidateID MaxCandCatID (and ultimately corresponds to category ID)
1 104 40
2 203 23
Then, re-joining back to your original table on these two inclusive of your AND CategoryID = 23 will only return the second CandidateID entry
And to help clarify to others who posted answers, the person does not appear to want the highest category ID, but if you look at them, they are sequentially added -- like an auto-incrementing number for the CandidateCategoryID. So, they want the most recent entry for a given candidate (hence candidates 1 & 2)... and if the last entry made was that of category = 23, they want THAT one.

Answer 2

select *
from (select t.*
from tbl t
join (select candidateid,
categoryid,
max(candidatecategoryid) as lastid
from tbl
group by candidateid, categoryid) v
on t.candidateid = v.candidateid
and t.categoryid = v.lastid) x
where categoryid = 23

Answer 3

This is a basic "greatest-of-n" problem. You can solve it with not exists, among other ways:
select t.*
from somedata t
where not exists (select 1
from somedata t2
where t2.categoryId = t.categoryId and
t2.candidateCategoryId > t.candidateCategoryId
);
EDIT:
If you only want categories where the max is 23, then add another condition:
select t.*
from somedata t
where not exists (select 1
from somedata t2
where t2.categoryId = t.categoryId and
t2.candidateCategoryId > t.candidateCategoryId
) and
t.categoryId = 23

Answer 4

Another way to skin this cat. Using your sample data, we can create an inline table for testing and get
DECLARE #candidates TABLE (CandidateCategoryId int,CandidateId int,CategoryId int)
INSERT INTO #candidates
SELECT 100, 1, 10
UNION
SELECT 101, 1, 11
UNION
SELECT 102, 1, 50
UNION
SELECT 103, 1, 23
UNION
SELECT 104, 1, 40
UNION
SELECT 200, 2, 20
UNION
SELECT 201, 2, 31
UNION
SELECT 202, 2, 12
UNION
SELECT 203, 2, 23
SELECT * FROM #candidates c
JOIN
(
SELECT CandidateId,MAX(CategoryId) AS CategoryId FROM #candidates
GROUP BY CandidateId
) tmp
ON c.CandidateId = tmp.CandidateId
AND c.CategoryId = tmp.CategoryId
And get results that look like
CandidateCategoryId | CandidateId | CategoryId
----------------------------------------------
201 | 2 | 31
102 | 1 | 50

Answer 5

I came up with this
select candidateCategoryId
from candidateCategory
where candidateCategoryId in (
select max(candidateCategoryId)
from candidateCategory
group by candidateId )
and categoryId = 23

Answer 6

select *
from yourtable
where candidateCategoryId = (select max(candidateCategoryId) from yourtable)

Answer 7

declare #categoryId int
select #categoryId = 23
with cte as
(
select candidateCategoryId, candidateId, categoryId,
rn = row_number() over (partition by candidateId order by candidateCategoryId desc)
from yourtable
)
select *
from cte c
where exists
(
select *
from cte x
where x.candidateId = c.candidateId
and x.rn = 1
and x.categoryId = #categoryId
)