sql query to find the Items with the highest difference - sql-server

Question

I have my database table ABC as shown below :
ItemId Month Year Sales
1 1 2013 333
1 2 2013 454
2 1 2013 434
and so on .
I would like to write a query to find the top 3 items that have had the highest increase in sales from last month to this month , so that I see somethinglike this in the output.
Output :
ItemId IncreaseInSales
1 +121
9 +33
6 +16
I came up to here :
select
(select Sum(Sales) from ABC where [MONTH] = 11 )
-
(select Sum(Sales) from ABC where [MONTH] = 10)
I cannot use a group by as it is giving an error . Can anyone point me how I can
proceed further ?

Answer 1

Assuming that you want the increase for a given month, you can also do this with an aggregation query:
select top 3 a.ItemId,
((sum(case when year = #YEAR and month = #MONTH then 1.0*sales end) /
sum(case when year = #YEAR and month = #MONTH - 1 or
year = #YEAR - 1 and #Month = 1 and month = 12
then sales end)
) - 1
) * 100 as pct_increase
from ABC a
group by a.ItemId
order by pct_increase desc;
You would put the year/month combination you care about in the variables #YEAR and #MONTH.
EDIT:
If you just want the increase, then do a difference:
select top 3 a.ItemId,
(sum(case when year = #YEAR and month = #MONTH then 1.0*sales end) -
sum(case when year = #YEAR and month = #MONTH - 1 or
year = #YEAR - 1 and #Month = 1 and month = 12
then sales
end)
) as difference
from ABC a
group by a.ItemId
order by difference desc;

Answer 2

Here is the SQL Fiddle that demonstrates the below query:
SELECT TOP(3) NewMonth.ItemId,
NewMonth.Month11Sales - OldMonth.Month10Sales AS IncreaseInSales
FROM
(
SELECT s1.ItemId, Sum(s1.Sales) AS Month11Sales
FROM ABC AS s1
WHERE s1.MONTH = 11
AND s1.YEAR = 2013
GROUP BY s1.ItemId
) AS NewMonth
INNER JOIN
(
SELECT s2.ItemId, Sum(s2.Sales) AS Month10Sales
FROM ABC AS s2
WHERE s2.MONTH = 10
AND s2.YEAR = 2013
GROUP BY s2.ItemId
) AS OldMonth
ON NewMonth.ItemId = OldMonth.ItemId
ORDER BY NewMonth.Month11Sales - OldMonth.Month10Sales DESC
You never mentioned if you could have more than one record for an ItemId with the same Month, so I made the query to handle it either way. Obviously you were lacking the year = 2013 in your query. Once you get past this year you will need that.

Answer 3

Another option could be something on these lines:
SELECT top 3 a.itemid, asales-bsales increase FROM
(
(select itemid, month, sum(sales) over(partition by itemid) asales from ABC where month=2
and year=2013) a
INNER JOIN
(select itemid, month, sum(sales) over(partition by itemid) bsales from ABC where month=1
and year=2013) b
ON a.itemid=b.itemid
)
ORDER BY increase desc
if you need to cater for months without sales then you can do a FULL JOIN and calculate increase as isnull(asales,0) - isnull(bsales,0)

Answer 4

You could adapt this solution based on PIVOT operator:
SET NOCOUNT ON;
DECLARE #Sales TABLE
(
ItemID INT NOT NULL,
SalesDate DATE NOT NULL,
Amount MONEY NOT NULL
);
INSERT #Sales (ItemID, SalesDate, Amount)
VALUES
(1, '2013-01-15', 333), (1, '2013-01-14', 111), (1, '2012-12-13', 100), (1, '2012-11-12', 150),
(2, '2013-01-11', 200), (2, '2012-12-10', 150), (3, '2013-01-09', 900);
-- Parameters (current year & month)
DECLARE #pYear SMALLINT = 2013,
#pMonth TINYINT = 1;
DECLARE #FirstDayOfCurrentMonth DATE = CONVERT(DATE, CONVERT(CHAR(4), #pYear) + '-' + CONVERT(CHAR(2), #pMonth) + '-01');
DECLARE #StartDate DATE = DATEADD(MONTH, -1, #FirstDayOfCurrentMonth), -- Begining of the previous month
#EndDate DATE = DATEADD(DAY, -1, DATEADD(MONTH, 1, #FirstDayOfCurrentMonth)) -- End of the current month
SELECT TOP(3) t.ItemID,
t.[2]-t.[1] AS IncreaseAmount
FROM
(
SELECT y.ItemID, y.Amount,
DENSE_RANK() OVER(ORDER BY y.FirstDayOfSalesMonth ASC) AS MonthNum -- 1=Previous Month, 2=Current Month
FROM
(
SELECT x.ItemID, x.Amount,
DATEADD(MONTH, DATEDIFF(MONTH, 0, x.SalesDate), 0) AS FirstDayOfSalesMonth
FROM #Sales x
WHERE x.SalesDate BETWEEN #StartDate AND #EndDate
) y
) z
PIVOT( SUM(z.Amount) FOR z.MonthNum IN ([1], [2]) ) t
ORDER BY IncreaseAmount DESC;
SQLFiddle demo

Answer 5

Your sample data seems to be incomplete, however, here is my try. I assume that you want to know the three items with the greatest sales-difference from one month to the next:
WITH Increases AS
(
SELECT a1.itemid,
a1.sales - (SELECT a2.sales
FROM dbo.abc a2
WHERE a1.itemid = a2.itemid
AND ( ( a1.year = a2.year
AND a1.month > 1
AND a1.month = a2.month + 1 )
OR ( a1.year = a2.year + 1
AND a1.month = 1
AND a2.month = 12 ) ))AS IncreaseInSales
FROM dbo.abc a1
)
SELECT TOP 3 ItemID, MAX(IncreaseInSales) AS IncreaseInSales
FROM Increases
GROUP BY ItemID
ORDER BY MAX(IncreaseInSales) DESC
Demo

Answer 6

SELECT
cur.[ItemId]
MAX(nxt.[Sales] - cur.[Sales]) AS [IncreaseInSales]
FROM ABC cur
INNER JOIN ABC nxt ON (
nxt.[Year] = cur.[Year] + cur.[month]/12 AND
nxt.[Month] = cur.[Month]%12 + 1
)
GROUP BY cur.[ItemId]

Answer 7

I'd do this this way. It should work in all the tagged versions of SQL Server:
SELECT TOP 3 [ItemId],
MAX(CASE WHEN [Month] = 2 THEN [Sales] END) -
MAX(CASE WHEN [Month] = 1 THEN [Sales] END) [Diff]
FROM t
WHERE [Month] IN (1, 2) AND [Year] = 2013
GROUP BY [ItemId]
HAVING COUNT(*) = 2
ORDER BY [Diff] DESC
Fiddle here.
The reason why I'm adding the HAVING clause is that if any item is added in only one of the months then the numbers will be all wrong. So I'm only comparing items that are only present in both months.
The reason of the WHERE clause would be to filter in advance only the needed months and improve the efficiency of the query.
An SQL Server 2012 solution could also be:
SELECT TOP 3 [ItemId], [Diff] FROM (
SELECT [ItemId],
LEAD([Sales]) OVER (PARTITION BY [ItemId] ORDER BY [Month]) - [Sales] Diff
FROM t
WHERE [Month] IN (1, 2) AND [Year] = 2013
) s
WHERE [Diff] IS NOT NULL
ORDER BY [Diff] DESC