I have the following table:
ClientId | CalculationDate | TransactedAmount |
1 13/02/2015 3
1 14/02/2015 3
2 14/02/2015 5
3 15/03/2015 6
2 15/03/2015 5
As a result I want table which contains ClientId and minimal Months passed since
maximum amount were transacted for each clientId.
How can i do that?
SELECT DISTINCT ClientId , MaxAmount.TransactedAmount , DATEDIFF(MONTH,MaxAmount.CalculationDate ,getdate())
FROM TABLENAME T1
CROSS APPLY(SELECT TOP 1 TransactedAmount ,CalculationDate
FROM TABLENAME T2
WHERE T1.ClientId = T2.ClientId
ORDER BY TransactedAmount DESC) MaxAmount
Using Window Function like Row_Nummber() we can achieve the same
;With cte(ClientId , CalculationDate , TransactedAmount )
AS
(
SELECT 1,'13/02/2015',3 UNION ALL
SELECT 1,'14/02/2015',3 UNION ALL
SELECT 2,'14/02/2015',5 UNION ALL
SELECT 3,'15/03/2015',6 UNION ALL
SELECT 2,'15/03/2015',5
)
SELECT ClientId , CalculationDate,TransactedAmount From
(
SELECT ClientId , CalculationDate ,MAX(TransactedAmount)OVER(Partition by ClientId Order by CalculationDate) As TransactedAmount ,
ROW_NUMBER()OVER(Partition by ClientId Order by CalculationDate) AS RNo From cte
)Dt
WHERE Dt.RNo=1
Related
I have bunch of data out of which I'm showing ID, max date and it's corresponding values (user id, type, ...). Then I need to take MAX date for each ID, substract 30 days and show first date and it's corresponding values within this date period.
Example:
ID Date Name
1 01.05.2018 AAA
1 21.04.2018 CCC
1 05.04.2018 BBB
1 28.03.2018 AAA
expected:
ID max_date max_name previous_date previous_name
1 01.05.2018 AAA 05.04.2018 BBB
I have working solution using subselects, but as I have quite huge WHERE part, refresh takes ages.
SUBSELECT looks like that:
(SELECT MIN(N.name)
FROM t1 N
WHERE N.ID = T.ID
AND (N.date < MAX(T.date) AND N.date >= (MAX(T.date)-30))
AND (...)) AS PreviousName
How'd you write the select?
I'm using TSQL
Thanks
I can do this with 2 CTEs to build up the dates and names.
SQL Fiddle
MS SQL Server 2017 Schema Setup:
CREATE TABLE t1 (ID int, theDate date, theName varchar(10)) ;
INSERT INTO t1 (ID, theDate, theName)
VALUES
( 1,'2018-05-01','AAA' )
, ( 1,'2018-04-21','CCC' )
, ( 1,'2018-04-05','BBB' )
, ( 1,'2018-03-27','AAA' )
, ( 2,'2018-05-02','AAA' )
, ( 2,'2018-05-21','CCC' )
, ( 2,'2018-03-03','BBB' )
, ( 2,'2018-01-20','AAA' )
;
Main Query:
;WITH cte1 AS (
SELECT t1.ID, t1.theDate, t1.theName
, DATEADD(day,-30,t1.theDate) AS dMinus30
, ROW_NUMBER() OVER (PARTITION BY t1.ID ORDER BY t1.theDate DESC) AS rn
FROM t1
)
, cte2 AS (
SELECT c2.ID, c2.theDate, c2.theName
, ROW_NUMBER() OVER (PARTITION BY c2.ID ORDER BY c2.theDate) AS rn
, COUNT(*) OVER (PARTITION BY c2.ID) AS theCount
FROM cte1
INNER JOIN cte1 c2 ON cte1.ID = c2.ID
AND c2.theDate >= cte1.dMinus30
WHERE cte1.rn = 1
GROUP BY c2.ID, c2.theDate, c2.theName
)
SELECT cte1.ID, cte1.theDate AS max_date, cte1.theName AS max_name
, cte2.theDate AS previous_date, cte2.theName AS previous_name
, cte2.theCount
FROM cte1
INNER JOIN cte2 ON cte1.ID = cte2.ID
AND cte2.rn=1
WHERE cte1.rn = 1
Results:
| ID | max_date | max_name | previous_date | previous_name |
|----|------------|----------|---------------|---------------|
| 1 | 2018-05-01 | AAA | 2018-04-05 | BBB |
| 2 | 2018-05-21 | CCC | 2018-05-02 | AAA |
cte1 builds the list of max_date and max_name grouped by the ID and then using a ROW_NUMBER() window function to sort the groups by the dates to get the most recent date. cte2 joins back to this list to get all dates within the last 30 days of cte1's max date. Then it does essentially the same thing to get the last date. Then the outer query joins those two results together to get the columns needed while only selecting the most and least recent rows from each respectively.
I'm not sure how well it will scale with your data, but using the CTEs should optimize pretty well.
EDIT: For the additional requirement, I just added in another COUNT() window function to cte2.
I would do:
select id,
max(case when seqnum = 1 then date end) as max_date,
max(case when seqnum = 1 then name end) as max_name,
max(case when seqnum = 2 then date end) as prev_date,
max(case when seqnum = 2 then name end) as prev_name,
from (select e.*, row_number() over (partition by id order by date desc) as seqnum
from example e
) e
group by id;
For example, i create a table about people contribue to 2 campaigns
+-------------------------------------+
| ID Name Campaign Amount (USD) |
+-------------------------------------+
| 1 A 1 10 |
| 2 B 1 5 |
| 3 C 2 7 |
| 4 D 2 9 |
+-------------------------------------+
Task: For each campaign, find the person (Name, ID) who contribute the most to
Expected result is
+-----------------------------------------+
| Campaign Name ID |
+-----------------------------------------+
| 1 A 1 |
| 2 D 4 |
+-----------------------------------------+
I used "group by Campaign" but the result have 2 columns "Campagin" and "max value" when I need "Name" and "ID"
Thanks for your help.
Edited: I fix some values, really sorry
You can use analytic functions for this:
select name, id, amount
from (select t.*, max(amount) over (partition by campaign) as max_amount
from t
) t
where amount = max_amount;
You can also do it by giving a rank/row_number partiton by campaign and order by descending order of amount.
Query
;with cte as(
select [num] = dense_rank() over(
partition by [Campaign]
order by [Amount] desc
), *
from [your_table_name]
)
select [Campaign], [Name], [ID]
from cte
where [num] = 1;
Try the next query:-
SELECT Campaign , Name , ID
FROM (
SELECT Campaign , Name , ID , MAX (Amount)
FROM MyTable
GROUP BY Campaign , Name , ID
) temp;
Simply use Where Clause with the max of amount group by Campaign:-
As following generic code:-
select a, b , c
from tablename
where d in
(
select max(d)
from tablename
group by a
)
Demo:-
Create table #MyTable (ID int , Name char(1), Campaign int , Amount int)
go
insert into #MyTable values (1,'A',1,10)
insert into #MyTable values (2,'B',1,5)
insert into #MyTable values (3,'C',2,7)
insert into #MyTable values (4,'D',2,9)
go
select Campaign, Name , ID
from #MyTable
where Amount in
(
select max(Amount)
from #MyTable
group by Campaign
)
drop table #MyTable
Result:-
Please find the below code for the same
SELECT *
FROM #MyTable T
OUTER APPLY (
SELECT COUNT(1) record
FROM #MyTable T1
where t.Campaign = t1.Campaign
and t.amount < t1.amount
)E
where E.record = 0
example table:
test_date | test_result | unique_ID
12/25/15 | 100 | 50
12/01/15 | 150 | 75
10/01/15 | 135 | 75
09/22/14 | 99 | 50
04/10/13 | 125 | 50
I need to find the first and last test date as well as the test result to match said date by user. So, I can group by ID, but not test result.
SELECT MAX(test_date)[need matching test_result],
MIN(test_date) [need matching test_result],
unique_id
from [table]
group by unique_id
THANKS!
Create TABLE #t
(
test_date date ,
Test_results int,
Unique_id int
)
INSERT INTO #t
VALUES ( '12/25/15',100,50 ),
( '12/01/15',150,75 ),
( '10/01/15',135,75 ),
( '09/22/14',99,50 ),
( '04/10/13',125,50 )
select 'MinTestDate' as Type, a.test_date, a.Test_results, a.Unique_id
from #t a inner join (
select min(test_date) as test_datemin, max(test_date) as test_datemax, unique_id from #t
group by unique_ID) b
on a.test_date = b.test_datemin
union all
select 'MaxTestDate' as Type, a.test_date, a.Test_results, a.Unique_id from #t a
inner join (
select min(test_date) as test_datemin, max(test_date) as test_datemax, unique_id from #t
group by unique_ID) b
on a.test_date = b.test_datemax
I would recommend window functions. The following returns the information on 2 rows per id:
select t.*
from (select t.*,
row_number() over (partition by unique_id order by test_date) as seqnum_asc,
row_number() over (partition by unique_id order by test_date desc) as seqnum_desc
from table t
) t;
For one row, use conditional aggregation (or pivot if you prefer):
select unique_id,
min(test_date), max(case when seqnum_asc = 1 then test_result end),
max(test_date), max(case when seqnum_desc = 1 then test_result end)
from (select t.*,
row_number() over (partition by unique_id order by test_date) as seqnum_asc,
row_number() over (partition by unique_id order by test_date desc) as seqnum_desc
from table t
) t
group by unique_id;
Consider using a combination of self-joins and derived tables:
SELECT t1.unique_id, minTable.MinOftest_date, t1.test_result As Mintestdate_result,
maxTable.MaxOftest_date, t2.test_result As Maxtestdate_result
FROM TestTable AS t1
INNER JOIN
(
SELECT Min(TestTable.test_date) AS MinOftest_date,
TestTable.unique_ID
FROM TestTable
GROUP BY TestTable.unique_ID
) As minTable
ON (t1.test_date = minTable.MinOftest_date
AND t1.unique_id = minTable.unique_id)
INNER JOIN TestTable As t2
INNER JOIN
(
SELECT Max(TestTable.test_date) AS MaxOftest_date,
TestTable.unique_ID
FROM TestTable
GROUP BY TestTable.unique_ID
) AS maxTable
ON t2.test_date = maxTable.MaxOftest_date
AND t2.unique_ID = maxTable.unique_ID
ON minTable.unique_id = maxTable.unique_id;
OUTPUT
unique_id MinOftest_date Mintestdate_result MaxOftest_date Maxtestdate_result
50 4/10/2013 125 12/25/2015 100
75 10/1/2015 135 12/1/2015 150
I am trying to get a specific row from a subquery, but I cannot use an aggregate function in a WHERE clause and I have read that I should be using a HAVING clause but I have no idea where to start.
This is my current sql statement:
SELECT *
FROM
(
select ID, SUM(BALANCE) AS Balance FROM bankacc GROUP BY ID
)A
I will get :
ID | Balance
1 | 30
2 | 40
3 | 50
4 | 50
I need the rows with the MAX(Balance), but I have no idea where to start, please help.
With window function:
DECLARE #t TABLE ( ID INT, Amount MONEY )
INSERT INTO #t
VALUES ( 1, 10 ),
( 1, 10 ),
( 1, 10 ),
( 2, 5 ),
( 2, 20 ),
( 3, 50 )
SELECT ID ,
Amount
FROM ( SELECT ID ,
SUM(Amount) AS Amount ,
RANK() OVER ( ORDER BY SUM(Amount) DESC ) AS rn
FROM #t
GROUP BY ID
) t
WHERE rn = 1
With TOP and TIES:
SELECT TOP 1 WITH TIES
ID ,
SUM(Amount) AS Amount
FROM #t
GROUP BY ID
ORDER BY Amount desc
These versions will return rows where sum will be max, not just top 1 row.
Output:
ID Amount
3 50.00
you can wrap it in a subquery:
SELECT q.id, max(q.b)
FROM
(
select ID, SUM(BALANCE) b FROM bankacc GROUP BY ID
) q
group by q.id
or order them in dessending order and get first record:
select top 1 ID, SUM(BALANCE) b FROM bankacc GROUP BY ID order by b desc
in MySQL you need to use limit 1 instead of top 1
I think this should be simple.
-- This will return only 1 record, even if there are 2 records for MAX same amount
SELECT top 1 ID ,
Amount
FROM ( SELECT ID ,
SUM(Amount) AS Amount
FROM Table
GROUP BY ID
) t
Order by Amount desc,ID asc
Using Window function : This will return what you want.
SELECT ID ,
Amount
FROM ( SELECT ID ,
SUM(Amount) AS Amount ,
RANK() OVER ( ORDER BY SUM(Amount) DESC ) AS rnk
FROM Table
GROUP BY ID
) t
WHERE rnk = 1
This is master table Query
Select *
from AC_TAB
where AC_ID = 7 ;
AC_PK AC_ID TYPE STATUS INS_DATE VALID
102 7 0 0 3/21/2012 3:35:08 PM 0
103 7 1 0 3/21/2012 3:35:08 PM
104 7 2 1 3/21/2012 3:35:08 PM
I am joining this table with txn table using ac_id. Since here its having 3 rows for ac_id 7 , my txn table returning 3 times. how to restrict this. since i want to return only one irrespective of type
MY Txn Query
Select txn_id, amount
from txn_hdr , ac_tab
where txn_ac_id = ac_id ;
txn_id amount
1 200
1 200
3 100
3 100
It is not actually clear what you need but it sounds like you only want to return one record per ac_id from the AC_TAB. If so, then there are a few ways that you can do this.
Using a subquery:
select *
from
(
select max(INS_DATE) INS_DATE, AC_ID
from ac_tab
group by AC_ID
) a
inner join txn_hdr t
on a.ac_id = t.ac_id;
Or with CTE using a row_number():
;with cte as
(
select a.ins_date, a.ac_id, t.amount, row_number()
over(partition by a.ac_id order by a.ins_date desc) rn
from ac_tab a
inner join txn_hdr t
on a.ac_id = t.ac_id
)
select *
from cte
where rn = 1;
Or with a row_number() in a subquery:
select *
from
(
select a.ins_date, a.ac_id, t.amount, row_number()
over(partition by a.ac_id order by a.ins_date desc) rn
from ac_tab a
inner join txn_hdr t
on a.ac_id = t.ac_id
) x
where rn = 1
You can do this way :
Select distinct txn_id, amount
from txn_hdr , ac_tab
where txn_ac_id = ac_id ;