This is my Data
Id Name Amt
1 ABC 20
2 XYZ 30
3 ABC 25
4 PQR 50
5 XYZ 75
6 PQR 40
I want the last record by every particular Name like :
3 ABC 25
5 XYZ 75
6 PQR 40
I tried group by, but i am missing some thing.
SELECT PatientID, Balance, PReceiptNo
FROM tblPayment
GROUP BY PatientID, Balance, PReceiptNo
Something like this should work:
SELECT p1.*
FROM tblPayment p1
LEFT JOIN tblPayment p2 ON p1.Name = p2.Name AND p1.Id < p2.Id
WHERE p2.Id IS NULL;
See this SQLFiddle
Should be similar to:
SELECT
id,
name,
amt
FROM
myTable mt1
where mt1.id = (
SELECT
MAX(id)
FROM myTable mt2
WHERE mt2.name = mt1.name
)
this should work
select * from (select id, name,amount from test order by id desc)as t1 group by name;
fiddle
One more option
SELECT *
FROM tblPayment p1
WHERE EXISTS (
SELECT 1
FROM tblPayment p2
WHERE p1.Name = p2.Name
HAVING MAX(p2.Id) = p1.Id
)
See demo on SQLFiddle
A possible solution:
Select p.*
from tblPayment p
Inner join (
Select name, max(id) as id
From tblPayment
Group by name
) as latest on latest.id = p.id
Related
I would like to get for each contract the record with the highest serial for the highest dates
ID CONTRCT C_DATE SERIAL
--------------------------------
1 ABC 20201201 1
2 ABC 20201201 2
3 ABC 20201201 3
4 DEF 20201201 3
4 DEF 20201210 1
5 DEF 20201210 2
Required results:
ID CONTRCT C_DATE SER
3 ABC 20201201 3
6 DEF 20201210 2
I achieved the results with two layers of self joins but may table is quite big and it takes a long time.
Is there a more efficient way?
My Query:
SELECT t3.ID
,t3.CONTRCT
,t3.C_DATE
,t3.SER
FROM (
SELECT ID
,tbl.CONTRCT
,tbl.C_DATE
,tbl.SER
FROM (
SELECT CONTRCT
,C_DATE
,max(SER) mx
FROM tbl
GROUP BY CONTRCT
,C_DATE
) t1
JOIN tbl ON t1.C_DATE = tbl.C_DATE
AND t1.mx = tbl.SER
AND t1.CONTRCT = tbl.CONTRCT
) t3
JOIN (
SELECT CONTRCT
,MAX(C_DATE) MAX_DATE
FROM (
SELECT ID
,tbl.CONTRCT
,tbl.C_DATE
,tbl.SER
FROM (
SELECT CONTRCT
,C_DATE
,max(SER) mx
FROM tbl
GROUP BY CONTRCT
,C_DATE
) t1
JOIN tbl ON t1.C_DATE = tbl.C_DATE
AND t1.mx = tbl.SER
AND t1.CONTRCT = tbl.CONTRCT
) t2
GROUP BY CONTRCT
) t4 ON t4.CONTRCT = t3.CONTRCT
AND t4.MAX_DATE = t3.C_DATE
I would suggest window functions:
select t.*
from (select t.*,
row_number() over (partition by contract order by date desc, ser desc) as seqnum
from tbl t
) t
where seqnum = 1;
It should works for you:
SELECT
t.contact,
MAX(t.C_DATE) C_DATE2,
(SELECT MAX(SERIAL) FROM test t2 WHERE t2.contact = t.contact AND t2.C_DATE=MAX(t.C_DATE) LIMIT 1) SERIAL
FROM test t
GROUP BY
t.contact;
If I was you, definitely will define an index of contact, date, serial on my table as well.
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.
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
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
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
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
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
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
I would like a result of
ID received total
2 25 25
from ItemReceived table
ID item received
2 1 5
2 2 2
2 1 10
2 2 8
and ItemsToReceive table
ID item quantity
2 1 15
2 2 10
Is there a way to get this result?
I used this code:
SELECT sum(ItemsToReceive.quantity), SUM( ItemReceived.received)
FROM ItemsToReceive
INNER JOIN ItemReceived ON ItemReceived.ID = ItemsToReceive.ID
GROUP BY poitems.poid
I got a wrong result making received to 50...
Can anyone help me?
Try this
select t1.Id,t1.quantity, t2.received from
(
SELECT ID,sum(quantity) quantity from ItemsToReceive group by id
) as t1 inner join
(
SELECT ID,SUM(received) as received from ItemReceived group by id
) as t2 on t1.id=t2.id
This might help
;WITH quantity AS (
SELECT sum(ItemsToReceive.quantity) As TotQuant
,ItemsToReceive.id AS id
FROM ItemsToReceive
GROUP BY ItemsToReceive.id
), Prod As (
SELECT SUM( ItemReceived.received) As TotRec
,ItemReceived.ID As Id
FROM ItemReceived
GROUP BY ItemReceived.ID
)
SELECT
q.Id
,p.TotRec As Received
,q.TotQuant As Quantity
FROM quantity q
INNER JOIN Prod p on p.Id = q.id
select ID , sum(received) , sum(quantity) from ItemReceived left join ItemsToReceive on (ItemReceived.ID = ItemsToReceive.ID and ItemReceived.item = ItemsToReceive.item) group by ID
I have a table_1:
id custno
1 1
2 2
3 3
and a table_2:
id custno qty descr
1 1 10 a
2 1 7 b
3 2 4 c
4 3 7 d
5 1 5 e
6 1 5 f
When I run this query to show the minimum order quantities from every customer:
SELECT DISTINCT table_1.custno,table_2.qty,table_2.descr
FROM table_1
LEFT OUTER JOIN table_2
ON table_1.custno = table_2.custno AND qty = (SELECT MIN(qty) FROM table_2
WHERE table_2.custno = table_1.custno )
Then I get this result:
custno qty descr
1 5 e
1 5 f
2 4 c
3 7 d
Customer 1 appears twice each time with the same minimum qty (& a different description) but I only want to see customer 1 appear once. I don't care if that is the record with 'e' as a description or 'f' as a description.
First of all... I'm not sure why you need to include table_1 in the queries to begin with:
select custno, min(qty) as min_qty
from table_2
group by custno;
But just in case there is other information that you need that wasn't included in the question:
select table_1.custno, ifnull(min(qty),0) as min_qty
from table_1
left outer join table_2
on table_1.custno = table_2.custno
group by table_1.custno;
"Generic" SQL way:
SELECT table_1.custno,table_2.qty,table_2.descr
FROM table_1, table_2
WHERE table_2.id = (SELECT TOP 1 id
FROM table_2
WHERE custno = table_1.custno
ORDER BY qty )
SQL 2008 way (probably faster):
SELECT custno, qty, descr
FROM
(SELECT
custno,
qty,
descr,
ROW_NUMBER() OVER (PARTITION BY custno ORDER BY qty) RowNum
FROM table_2
) A
WHERE RowNum = 1
If you use SQL-Server you could use ROW_NUMBER and a CTE:
WITH CTE AS
(
SELECT table_1.custno,table_2.qty,table_2.descr,
RN = ROW_NUMBER() OVER ( PARTITION BY table_1.custno
Order By table_2.qty ASC)
FROM table_1
LEFT OUTER JOIN table_2
ON table_1.custno = table_2.custno
)
SELECT custno, qty,descr
FROM CTE
WHERE RN = 1
Demolink
I have this table:
supplier | product | qty
--------------------------
s1 | p1 | 300
s1 | p2 | 90
s2 | p3 | 89
I want to find suppliers with more than 2 products.
But only with select and where, no group by. Any suggestion?
Why would you want not to use group by is beyond me, but this might work:
SELECT Supplier FROM table outer WHERE
(
select count(Products) from table inner
where inner.Supplier = outer.Supplier
) > 2
Please bear in mind, that group by is made for stuff like that and should be used.
;WITH
sequenced_data AS
(
SELECT
supplier,
ROW_NUMBER() OVER (PARTITION BY supplier ORDER BY product) AS supplier_product_ordinal
FROM
YourTable
)
SELECT
supplier
FROM
sequenced_data
WHERE
supplier_product_ordinal = 3
But I'd expect it to be slower than using GROUP BY.
SELECT DISTINCT
supplier
FROM
yourTable
WHERE
EXISTS (SELECT * FROM yourTable AS lookup WHERE supplier = yourTable.supplier AND product < yourTable.product)
AND EXISTS (SELECT * FROM yourTable AS lookup WHERE supplier = yourTable.supplier AND product > yourTable.product);
In the usual parts and suppliers database, this relvar is named SP:
SELECT DISTINCT T1.SNO
FROM SP AS T1
JOIN SP AS T2
ON T1.SNO = T2.SNO
AND T2.PNO <> T1.PNO
JOIN SP AS T3
ON T1.SNO = T3.SNO
AND T3.PNO <> T1.PNO
AND T3.PNO <> T2.PNO;
Noting that you can use HAVING without GROUP BY:
SELECT DISTINCT T1.SNO
FROM SP AS T1
WHERE EXISTS (
SELECT 1
FROM SP AS T2
WHERE T2.SNO = T1.SNO
HAVING COUNT(*) > 2
);
;WITH T AS
(
SELECT *,
COUNT(*) OVER (PARTITION BY S) AS Cnt
FROM YourTable
)
SELECT DISTINCT S
FROM T
WHERE Cnt > 2
with subquery:
select distinct supplier
from table a
where (select count(*)
from table b
where b.supplier = a.supplier and b.product <> a.product
) > 1