sql joining table, not providing perfect result - sql

Question

let me explain what i need, i have 2 table named A and B. B is sub table for A.
Here is Schema:
------------------------
Table B:
itemId version qty AId
44 1 1 200
44 1 2 201
44 2 2 200
------------------------
Table A:
id tId
200 100
201 100
------------------------
and here is what i need: i need sum of all latest version qty that have same tId.
here is my query:
select sum(qty) as sum from B
left join A on A.id=B.AId
where itemId=44 and tId=100 and
version=(select max(version) from B where itemId=44 and tId=100)
the result get wrong when one item got version 2 and version 1 ignored.
thanks.
EDIT:
what exactly i need is:
itemId version qty AId
44 2 2 200
44 1 2 201
And Result of Sum(qty) must be 4, because they have same tId and they have Max version in each AId.

Answer 1

Use window function.
select itemid, version, qty, aid
from (
select *, max(version) over (partition by AId) as latestVersion
from B
) as B
where version = latestVersion
to sum up
select tId, SUM(qty) AS qty_sum
from (
select *, max(version) over (partition by AId) as latestVersion
from B
) as B
join A on B.AId = A.id
where version = latestVersion
group by tId

Answer 2

Working solution
select b.* from B as b
inner join
(select AID,itemId,Max(version) as mVersion from B
group by AID,itemID) d
on b.AID = d.AID and b.itemID = d.itemID and b.Version = d.mVersion
inner join A as a
on B.AID = a.id
where b.itemID = 44 --apply if you need
result
itemid version qty aid
44 2 2 200
44 1 2 201
this will give you result as you sum of quantity
select itemID,sum(qty) from (
select b.* from B as b
inner join
(select AID,itemId,Max(version) as mVersion from B
group by AID,itemID) d
on b.AID = d.AID and b.itemID = d.itemID and b.Version = d.mVersion
inner join A as a
on B.AID = a.id
where b.itemID = 44 --apply if you need
) e group by itemID
result
itemid sum
44 4

Answer 3

Try This one
DECLARE #TA Table (id int,tid int)
DECLARE #TB Table (itemid int, version int,qty int,AID int)
INSERT INTO #TA
SELECT 200, 100
UNION ALL
SELECT 201, 100
INSERT INTO #TB
SELECT 44,1,1,201
UNION ALL
SELECT 44,1,2,200
UNION ALL
SELECT 44,2,3,200
UNION ALL
SELECT 44,2,5,201
DECLARE #tid int
SET #tid = 100
SELECT XB.* FROM #Tb XB INNER JOIN
(SELECT Version,Max(AID) Aid FROM #TA A INNER JOIN #TB B ON A.id = B.AID AND tid = #tid Group By Version) X
ON X.version = XB.version and XB.AID = X.Aid

Answer 4

i think this query help you to solve your problem
SELECT itemId, version, qty , AId FROM (
SELECT itemId, version, qty , AId FROM b
LEFT JOIN a ON (b.aid = a.id)
) temp
WHERE version = (SELECT MAX(version) FROM b WHERE b.aid = temp.aid)
and temp.tid = 100 and temp.itemId = 44

Answer 5

SELECT B.*
FROM B
INNER JOIN
(SELECT Aid,MAX(version) AS version FROM B WHERE itemId=44 GROUP BY AId) AS B1
ON B.Aid=B1.Aid
AND B.version=B1.version
INNER JOIN
(SELECT * FROM A WHERE tId=100) AS A
ON A.id=B.Aid
Order BY B.aid
For Sum of qty
SELECT SUM(B.qty)
FROM B
INNER JOIN
(SELECT Aid,MAX(version) AS version FROM B WHERE itemId=44 GROUP BY AId) AS B1
ON B.Aid=B1.Aid AND B.version=B1.version
INNER JOIN
(SELECT * FROM A WHERE tId=100) AS A
ON A.id=B.Aid
GROUP BY A.tid
Output
itemid version qty aid
44 2 2 200
44 1 2 201
Demo
http://sqlfiddle.com/#!17/092dd/5

Answer 6

The most efficient solution greatest-n-per-group problems in Postgres are typically using the (proprietary) operator distinct on ()
So to get the latest version for each a.id, you can use:
select distinct on (a.id) b.*
from a
join b on a.id = b.aid
order by a.id, b.version desc;
The above returns:
itemid | version | qty | aid
-------+---------+-----+----
44 | 2 | 2 | 200
44 | 1 | 2 | 201
You can then sum over the result:
select sum(qty)
from (
select distinct on (a.id) b.qty
from a
join b on a.id = b.aid
order by a.id, b.version desc
) t;
Note that normally an order by in a derived table is useless, but in this case it's needed because otherwise distinct on () wouldn't work.
Online example: http://rextester.com/DRHK19268