Could anyone suggest the query for the below scenario?
Table is as below
MachineName ManufacturedBy Amount
---------------------------------
A X 50
B X 50
C Q 30
D Q 30
The data should be as follows
MachineName ManufacturedBy Amount
----------------------------------
A X 50
B X 50
Subtotal 100
C Q 30
D Q 30
Subtotal 60
Grandtotal
Thanks
Sasi
You may use GROUP BY with ROLLUP:
SELECT
COALESCE(ManufacturedBy, 'All Manufacturers') AS ManufacturedBy,
COALESCE(MachineName, 'All Machines') AS MachineName,
SUM(Amount) AS Amount
FROM yourTable
GROUP BY ROLLUP (ManufacturedBy, MachineName);
Demo
I tried the below. It also worked for me.
SELECT
ManufacturedBy ,
MachineName ,
SUM AS Amount
FROM yourTable
GROUP BY GROUPING SETS (ManufacturedBy,(ManufacturedBy, MachineName),());
Thanks
In T-SQL (MSSql 2008R2) I Would like to select certain rows from a table or set of results;
StoreId StoreName BrochureId PageId Rank Distance
43561 X 1627 11608 73 598.10
43561 X 1627 11591 68 598.10
43561 X 1627 11615 41 598.10
43827 Y 1727 21708 75 1414.69
43827 Y 1727 21591 62 1414.69
43827 Y 1727 21615 44 1414.69
43919 Z 1827 31809 77 2487.35
43919 Z 1827 31591 60 2487.35
43919 Z 1827 31615 39 2487.35
Would like to select only rows with lowest distance and with the highest rank, as such;
StoreId StoreName BrochureId PageId Rank Distance
43561 X 1627 11608 73 598.10
43827 Y 1727 21708 75 1414.69
43919 Z 1827 31809 77 2487.35
Thank you for your help.
You can use ROW_NUMBER for this.
SELECT * FROM (
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY StoreName ORDER BY Distance, [Rank] DESC) Rn
FROM
Table1
) t
WHERE Rn = 1
there are other ranking functions you can use as well.. for example, if you use RANK instead of ROW_NUMBER here, you can include ties in your result as well.
SQL Ranking Functions
Use a correlated subquery in your where clause... assuming your table name is MyTable, something like this should get what you want:
SELECT [StoreId], [StoreName], [BrochureId], [PageId], [Rank], [Distance]
FROM MyTable m
WHERE [Rank] = (SELECT MAX([Rank]) FROM MyTable x WHERE x.StoreId = m.StoreId)
OR [Distance] = (SELECT MIN([Distance]) FROM MyTable y WHERE y.StoreId = m.StoreId)
(note, I enclosed the column names in square brackets because "Rank" is a reserved SQL Keyword)
This is the input table:
Customer_ID Date Amount
1 4/11/2014 20
1 4/13/2014 10
1 4/14/2014 30
1 4/18/2014 25
2 5/15/2014 15
2 6/21/2014 25
2 6/22/2014 35
2 6/23/2014 10
There is information pertaining to multiple customers and I want to get a rolling sum across a 3 day window for each customer.
The solution should be as below:
Customer_ID Date Amount Rolling_3_Day_Sum
1 4/11/2014 20 20
1 4/13/2014 10 30
1 4/14/2014 30 40
1 4/18/2014 25 25
2 5/15/2014 15 15
2 6/21/2014 25 25
2 6/22/2014 35 60
2 6/23/2014 10 70
The biggest issue is that I don't have transactions for each day because of which the partition by row number doesn't work.
The closest example I found on SO was:
SQL Query for 7 Day Rolling Average in SQL Server
but even in that case there were transactions made everyday which accomodated the rownumber() based solutions
The rownumber query is as follows:
select customer_id, Date, Amount,
Rolling_3_day_sum = CASE WHEN ROW_NUMBER() OVER (partition by customer_id ORDER BY Date) > 2
THEN SUM(Amount) OVER (partition by customer_id ORDER BY Date ROWS BETWEEN 2 PRECEDING AND CURRENT ROW)
END
from #tmp_taml9
order by customer_id
I was wondering if there is way to replace "BETWEEN 2 PRECEDING AND CURRENT ROW" by "BETWEEN [DATE - 2] and [DATE]"
One option would be to use a calendar table (or something similar) to get the complete range of dates and left join your table with that and use the row_number based solution.
Another option that might work (not sure about performance) would be to use an apply query like this:
select customer_id, Date, Amount, coalesce(Rolling_3_day_sum, Amount) Rolling_3_day_sum
from #tmp_taml9 t1
cross apply (
select sum(amount) Rolling_3_day_sum
from #tmp_taml9
where Customer_ID = t1.Customer_ID
and datediff(day, date, t1.date) <= 3
and t1.Date >= date
) o
order by customer_id;
I suspect performance might not be great though.
I have some data in a table as follows:
FileDate SumAmount
20150401 90.99
20150401 313
20150403 481.2
20150404 321.27
20150405 103
20150406 25
20150407 180.5
20150408 319.91
20150409 688
20150411 69
20150412 65
20150413 322
20150414 100
20150415 111.97
20150416 979.15
20150417 655.4
20150418 124
20150419 30
20150420 457
20150421 192.6
20150422 191.96
20150423 220
20150424 252.5
20150425 109.1
20150426 135.25
20150427 648.08
20150428 692
20150429 410.99
20150430 170
20150501 166.19
20150502 92
20150503 100
20150504 59
20150505 124.01
20150506 44.5
20150507 331.64
20150508 299.8
I am trying to devise a query that will find the highest 4 consecutive days values in the data.
Essentially, I think I need to partition by date and perform a row numbering over it but I can't seem to get the syntax right to evaluate the values.
So I use -3 in the join conditions since the day itself counts as one. Let me know what you think. Also I use day of year(DY) to ensure that it's only consecutive days and so I don't have to rank the dates manually. Hope this helps!
DECLARE #yourTable TABLE(FileDate DATE ,SumAmount FLOAT);
INSERT INTO #yourTable
VALUES ('20150401',90.99),
('20150402',313),
('20150403',481.2),
('20150404',321.27),
('20150405',103),
('20150406',25),
('20150407',180.5),
('20150408',319.91),
('20150409',688),
('20150411',69),
('20150412',65),
('20150413',322),
('20150414',100),
('20150415',111.97),
('20150416',979.15),
('20150417',655.4),
('20150418',124),
('20150419',30),
('20150420',457),
('20150421',192.6),
('20150422',191.96),
('20150423',220),
('20150424',252.5),
('20150425',109.1),
('20150426',135.25),
('20150427',648.08),
('20150428',692),
('20150429',410.99),
('20150430',170),
('20150501',166.19),
('20150502',92),
('20150503',100),
('20150504',59),
('20150505',124.01),
('20150506',44.5),
('20150507',331.64),
('20150508',299.8);
WITH CTE
AS
(
SELECT YEAR(FileDate) yr,DATEPART(DY,FileDate) dy,fileDate,SumAmount
FROM #yourTable
),
CTE_Max_Sum
AS
(
SELECT TOP 1 A.yr,A.dy,A.FileDate,SUM(B.SumAmount) consec4DaySum
FROM CTE A
INNER JOIN CTE B
ON B.dy BETWEEN A.dy - 3 AND A.dy
AND A.yr = B.yr
GROUP BY A.yr,A.dy,A.FileDate
ORDER BY SUM(B.SumAmount) DESC
)
SELECT A.*,B.consec4DaySum
FROM CTE A
INNER JOIN CTE_Max_Sum B
ON A.dy BETWEEN B.dy - 3 AND B.dy
AND A.yr = B.yr
Results:
yr dy fileDate SumAmount consec4DaySum
----------- ----------- ---------- ---------------------- ----------------------
2015 117 2015-04-27 648.08 1921.07
2015 118 2015-04-28 692 1921.07
2015 119 2015-04-29 410.99 1921.07
2015 120 2015-04-30 170 1921.07
You can use a CTE for that, joining every row with its three following rows (day-wise) and summing up. This Fiddle sadly does not work for me, it runs on my sql server and work for you. Watch out for recursion depth, without WHERE cte.Consecutive < 4 you quickly run into an error.
WITH cte (StartDate, EndDate, Consecutive, SumAmount)
AS (
SELECT t.FileDate, t.FileDate, 1, t.SumAmount FROM dbo.table30194903 t
UNION ALL
SELECT cte.StartDate, t.FileDate, cte.Consecutive + 1, cte.SumAmount + t.SumAmount
FROM dbo.table30194903 t INNER JOIN cte ON DATEADD(DAY, 1, cte.EndDate) = t.FileDate
WHERE cte.Consecutive < 5
)
SELECT *
FROM cte
WHERE cte.Consecutive = 4
ORDER BY cte.SumAmount DESC
EDIT: Had two errors in my query, it summed up wrong rows and showd the last day in the series.
I would like to add an answer using a subquery, however it does take more time compared to my cte...
SELECT t.FileDate, SUM(s.SumAmount)
FROM dbo.table30194903 t
LEFT JOIN dbo.table30194903 s ON t.FileDate <= s.FileDate AND DATEDIFF(DAY, t.FileDate, s.FileDate) < 4
GROUP BY t.FileDate
HAVING COUNT(s.SumAmount) = 4
ORDER BY SUM(s.SumAmount) DESC
I think the simplest way to get this is to use an APPLY to get the number of records in the n days following each row, and then limit this to where there are n dates, this ensures you have consecutive days. You can then just order by the sum and select the top 1:
DECLARE #n INT = 4;
SELECT TOP 1
FirstDate = t.FileDate,
FourDaySum = t2.Amount
FROM dbo.T
CROSS APPLY
( SELECT Amount = SUM(t2.SumAmount),
Dates = COUNT(DISTINCT t2.FileDate)
FROM dbo.T AS t2
WHERE t2.FileDate >= t.FileDate
AND t2.FileDate < DATEADD(DAY, #n, t.FileDate)
) AS t2
WHERE t2.Dates = #n
ORDER BY t2.Amount DESC;
Example on SQL Fiddle
How about a simply while block and sum the values of a range of dates?
DECLARE #startingDate DATETIME, #searchDate DATETIME;
DECLARE #maxSoFar INT, #sum INT, #daysRange INT;
SET #startingDate = convert(datetime, '20150401', 110)
SET #searchDate = #startingDate;
SET #daysRange = 3;
SET #maxSoFar = 0;
WHILE GETDATE()> #searchDate
BEGIN
--PRINT #searchDate
--PRINT DATEADD(DAY,#daysRange,#searchDate)
SELECT #sum = SUM(SumAmount) FROM MyTable WHERE FileDate >= #searchDate AND FileDate <= DATEADD(DAY,#daysRange,#searchDate)
IF #sum > #maxSoFar
BEGIN
SET #maxSoFar = #sum;
END
SET #searchDate = DATEADD(DAY,1,#searchDate)
END
Say I have a table like the following:
PK Code Value
1 A 200
2 A 300
3 A 25
4 A 75
5 A 50
6 A 15
7 A 300
8 A 75
How would I get the value of the top 4 highest values where code=A (i.e. just want the sum of 300 + 300 + 200 + 75)
Thanks
You can use a derived table or Common Table Expression to get the top 4 then SUM that.
SELECT SUM(Value) As Top4Sum
FROM
(
SELECT TOP (4) Value
FROM YourTable
WHERE Code = 'A'
ORDER BY Value DESC
) T
If you wanted the SUM of the TOP 4 for every Code you could do
;WITH CTE
AS (SELECT *,
ROW_NUMBER() OVER (PARTITION BY Code ORDER BY Value DESC) RN
FROM YourTable)
SELECT Code,
SUM(Value)
FROM CTE
WHERE RN <= 4
GROUP BY Code