This question has been covered similarly before BUT I'm struggling.
I need to find top N sales based on customer buying patterns..
ideally this needs to be top N by customer by Month Period by Year but for now i'm just looking at top N over the whole DB.
My query looks like:
-- QUERY TO SHOW TOP 2 CUSTOMER INVOICES BY CUSTOMER BY MONTH
SELECT
bill_to_code,
INVOICE_NUMBER,
SUM( INVOICE_AMOUNT_CORP ) AS 'SALES',
ROW_NUMBER() OVER ( PARTITION BY bill_to_code ORDER BY SUM( INVOICE_AMOUNT_CORP ) DESC ) AS 'Row'
FROM
FACT_OM_INVOICE
JOIN dim_customer_bill_to ON FACT_OM_INVOICE.dim_customer_bill_to_key = dim_customer_bill_to.dim_customer_bill_to_key
--WHERE
-- 'ROW' < 2
GROUP BY
invoice_number,
Dim_customer_bill_to.bill_to_code
I can't understand the solutions given to restrict Row to =< N.
Please help.
Try this.
-- QUERY TO SHOW TOP 2 CUSTOMER INVOICES BY CUSTOMER BY MONTH
;WITH Top2Customers
AS
(
SELECT
bill_to_code,
INVOICE_NUMBER,
SUM( INVOICE_AMOUNT_CORP ) AS 'SALES',
ROW_NUMBER() OVER ( PARTITION BY bill_to_code ORDER BY SUM( INVOICE_AMOUNT_CORP ) DESC )
AS 'RowNumber'
FROM
FACT_OM_INVOICE
JOIN dim_customer_bill_to ON FACT_OM_INVOICE.dim_customer_bill_to_key = dim_customer_bill_to.dim_customer_bill_to_key
GROUP BY
invoice_number,
Dim_customer_bill_to.bill_to_code
)
SELECT * FROM Top2Customers WHERE RowNumber < 3
You have to wrap your select into another to use the value produced by row_number()
select * from (
SELECT
bill_to_code,
INVOICE_NUMBER,
SUM( INVOICE_AMOUNT_CORP ) AS SALES,
ROW_NUMBER() OVER ( PARTITION BY bill_to_code ORDER BY SUM( INVOICE_AMOUNT_CORP ) DESC ) AS RowNo
FROM
FACT_OM_INVOICE
JOIN dim_customer_bill_to ON FACT_OM_INVOICE.dim_customer_bill_to_key = dim_customer_bill_to.dim_customer_bill_to_key
--WHERE
-- 'ROW' < 2
GROUP BY
invoice_number,
Dim_customer_bill_to.bill_to_code
) base where RowNo < 2
Related
I have some prices for the month of January.
Date,Price
1,100
2,100
3,115
4,120
5,120
6,100
7,100
8,120
9,120
10,120
Now, the o/p I need is a non-overlapping date range for each price.
price,from,To
100,1,2
115,3,3
120,4,5
100,6,7
120,8,10
I need to do this using SQL only.
For now, if I simply group by and take min and max dates, I get the below, which is an overlapping range:
price,from,to
100,1,7
115,3,3
120,4,10
This is a gaps-and-islands problem. The simplest solution is the difference of row numbers:
select price, min(date), max(date)
from (select t.*,
row_number() over (order by date) as seqnum,
row_number() over (partition by price, order by date) as seqnum2
from t
) t
group by price, (seqnum - seqnum2)
order by min(date);
Why this works is a little hard to explain. But if you look at the results of the subquery, you will see how the adjacent rows are identified by the difference in the two values.
SELECT Lag.price,Lag.[date] AS [From], MIN(Lead.[date]-Lag.[date])+Lag.[date] AS [to]
FROM
(
SELECT [date],[Price]
FROM
(
SELECT [date],[Price],LAG(Price) OVER (ORDER BY DATE,Price) AS LagID FROM #table1 A
)B
WHERE CASE WHEN Price <> ISNULL(LagID,1) THEN 1 ELSE 0 END = 1
)Lag
JOIN
(
SELECT [date],[Price]
FROM
(
SELECT [date],Price,LEAD(Price) OVER (ORDER BY DATE,Price) AS LeadID FROM [#table1] A
)B
WHERE CASE WHEN Price <> ISNULL(LeadID,1) THEN 1 ELSE 0 END = 1
)Lead
ON Lag.[Price] = Lead.[Price]
WHERE Lead.[date]-Lag.[date] >= 0
GROUP BY Lag.[date],Lag.[price]
ORDER BY Lag.[date]
Another method using ROWS UNBOUNDED PRECEDING
SELECT price, MIN([date]) AS [from], [end_date] AS [To]
FROM
(
SELECT *, MIN([abc]) OVER (ORDER BY DATE DESC ROWS UNBOUNDED PRECEDING ) end_date
FROM
(
SELECT *, CASE WHEN price = next_price THEN NULL ELSE DATE END AS abc
FROM
(
SELECT a.* , b.[date] AS next_date, b.price AS next_price
FROM #table1 a
LEFT JOIN #table1 b
ON a.[date] = b.[date]-1
)AA
)BB
)CC
GROUP BY price, end_date
How can I find the sales margin every Day via SQL, assuming they are sold in the order they were purchased?
Please try this solution -
;with cte as
(
select purchase_date,item,cost, qty as num from purchase
union all
select purchase_date,item,cost, num-1 from cte where num>1
),
cte2 as
(
select sale_date,item,price, qty as num from sales
union all
select sale_date,item,price, num-1 from cte2 where num>1
)
select sale_date, sum(price-cost) from (
(select sale_date, item, price ,row_number() over (order by sale_date,num) rn from cte2) s
inner join
(select purchase_date, item, cost ,row_number() over (order by purchase_date,num) rn2 from cte) z
on s.item=z.item and s.rn=z.rn2)
group by sale_date
The following query displays duplicates in a table with the qty alias showing the total count, eg if there are five duplicates then all five will have the same qty = 5.
select s.*, t.*
from [Migrate].[dbo].[Table1] s
join (
select [date] as d1, [product] as h1, count(*) as qty
from [Migrate].[dbo].[Table1]
group by [date], [product]
having count(*) > 1
) t on s.[date] = t.[d1] and s.[product] = t.[h1]
ORDER BY s.[product], s.[date], s.[id]
Is it possible to amend the count(*) as qty to show an incremental count so that five duplicates would display 1,2,3,4,5?
The answer to your question is row_number(). How you use it is rather unclear, because you provide no guidance, such as sample data or desired results. Hence this answer is rather general:
select s.*, t.*,
row_number() over (partition by s.product order by s.date) as seqnum
from [Migrate].[dbo].[Table1] s join
(select [date] as d1, [product] as h1, count(*) as qty
from [Migrate].[dbo].[Table1]
group by [date], [product]
having count(*) > 1
) t
on s.[date] = t.[d1] and s.[product] = t.[h1]
order by s.[product], s.[date], s.[id];
The speculation is that the duplicates are by product. This enumerates them by date. Some combination of the partition by and group by is almost certainly what you need.
I have a date column Order_date and I am looking for ways to calculate the date difference between customer last order date and his recent previous ( previous form last) order_date ....
Example
Customer : 1, 2 , 1 , 1
Order_date: 01/02/2007, 02/01/2015, 06/02/2014, 04/02/2015
As you can see customer # 1 has three orders.
I want to know the date difference between his recent order date (04/02/2015) and his recent previous (06/02/2014).
For SQL Server 2012 & 2014 you could use LAG with a DATEDIFF to see the number of days between them.
For older versions, a CTE would probably be your best bet:
;WITH CTE AS
(
SELECT CustomerID,
Order_Date,
rn = ROW_NUMBER() OVER (PARTITION BY CustomerID ORDER BY Order_Date DESC)
)
SELECT c1.CustomerID,
DATEDIFF(d, c1.Order_Date, c2.Order_Date)
FROM CTE c1
INNER JOIN CTE c2 ON c2.rn = c1.rn + 1
In SQL Server 2012+, you can use lag() to get the difference between any two dates:
select t.*,
datediff(day, lag(order_date) over (partition by customer order by order_date),
order_date) as days_dff
from table t;
If you have an older version, you can do something similar with correlated subqueries or outer apply.
EDIT:
If you just want the difference between the two most recent dates, use conditional aggregation instead:
select customer,
datediff(day, max(case when seqnum = 2 then order_date end),
max(case when seqnum = 1 then order_date end)
) as MostRecentDiff
from (select t.*,
row_number() over (partition by customer order by order_date desc) as seqnum
from table t
) t
group by customer;
If you're using SQL Server 2008 or later, you can try CROSS APPLY.
SELECT [customers].[customer_id], DATEDIFF(DAY, MIN([recent_orders].[order_date]), MAX([recent_orders].[order_date])) AS [elapsed]
FROM [customers]
CROSS APPLY (
SELECT TOP 2 [order_date]
FROM [orders]
WHERE ([orders].[customer_id] = [customers].[customer_id])
) [recent_orders]
GROUP BY [customers].[customer_id]
SELECT DATEDIFF(DAY, Y.PrevLastOrderDate, Y.LastOrderDate) AS PreviousDays
FROM
(
SELECT X.LastOrderDate
, (SELECT MAX(OrderDate) FROM dbo.Orders SO WHERE SO.CustomerID=1 AND SO.OrderDate < X.LastOrderDate) AS PrevLastOrderDate
FROM
(
select MAX(OrderDate) AS LastOrderDate
FROM dbo.Orders O
WHERE O.CustomerID=1
)X
)Y
drop table #Invoices
create table #Invoices ( OrderId int , OrderDate datetime )
insert into #Invoices (OrderId , OrderDate )
select 101, '01/01/2001' UNION ALL Select 202, '02/02/2002' UNION ALL Select 303, '03/03/2003'
UNION ALL Select 808, '08/08/2008' UNION ALL Select 909, '09/09/2009'
;
WITH
MyCTE /* http://technet.microsoft.com/en-us/library/ms175972.aspx */
( OrderId,OrderDate,ROWID) AS
(
SELECT
OrderId,OrderDate
, ROW_NUMBER() OVER ( ORDER BY OrderDate ) as ROWID
FROM
#Invoices inv
)
SELECT
OrderId,OrderDate
,(Select Max(OrderDate) from MyCTE innerAlias where innerAlias.ROWID = (outerAlias.ROWID-1) ) as PreviousOrderDate
,
[MyDiff] =
CASE
WHEN (Select Max(OrderDate) from MyCTE innerAlias where innerAlias.ROWID = (outerAlias.ROWID-1) ) iS NULL then 0
ELSE DATEDIFF (mm, OrderDate , (Select Max(OrderDate) from MyCTE innerAlias where innerAlias.ROWID = (outerAlias.ROWID-1) ) )
END
, ROWIDMINUSONE = (ROWID-1)
, ROWID as ROWID_SHOWN_FOR_KICKS , OrderDate as OrderDateASecondTimeForConvenience
FROM
MyCTE outerAlias
ORDER BY outerAlias.OrderDate Desc , OrderId
Declare #sec_temp table
(
sec_no varchar(10),
amount money,
price_date date
)
insert #sec_temp
values
('123ABC', 25, '2011-01-20'),
('123ABC', 25, '2011-01-19'),
('123ABC', 25, '2011-01-18'),
('123ABC', 20, '2011-01-15'),
('123ABC', 22, '2011-01-13'),
('456DEF', 22, '2011-01-13')
Problem: To list out the distinct sec_no with the latest price (amount) and the number of days it was at the current price. In this case,
Result:
sec_no amount no_of_days_at_price
123ABC 25 3 e.g. 01-18 to 01-20
456DEF 22 1 e.g. 01-13
select
a.sec_no,
a.amount,
min(price_date) as FirstDateAtPrice,
No_of_days_at_price = COALESCE(DATEDIFF(d, c.price_date, a.price_date),0)
from (
select *, ROW_NUMBER() over (partition by sec_no order by price_date desc) rn
from #sec_temp) a
outer apply (
select top 1 *
from #sec_temp b
where a.sec_no=b.sec_no and a.amount <> b.amount
order by b.price_date desc
) c
where a.rn=1
The subquery A works out the greatest-1-per-group, which is to say the most recent price record for each sec_no. The subquery C finds the first prior record that holds a different price for the same sec_no. The difference in the two dates is the number of days sought. If you need it to be one for no prior date, change the end of the COALESCE line to 1 instead of 0.
EDITED for clarified question
To start counting from the first date equal to the current rate, use this query instead
select
sec_no,
amount,
No_of_days_at_price = 1 + DATEDIFF(d, min(price_date), max(price_date))
from (
select *,
ROW_NUMBER() over (partition by sec_no order by price_date desc) rn,
ROW_NUMBER() over (partition by sec_no, amount order by price_date desc) rn2
from #sec_temp
) X
WHERE rn=rn2
group by sec_no, amount
AND FINALLY If the required result is actually the days between
the first date on which the price is equal to current; and
today
Then the only part to change is this:
No_of_days_at_price = 1 + DATEDIFF(d, min(price_date), getdate())
Here's one approach, first looking up the latest price, and then the last price that was different:
select secs.sec_no
, latest.amount as price
, case when previous.price_date is null then 1
else datediff(day, previous.price_date, latest.price_date)
end as days_at_price
from (
select distinct sec_no
from #sec_temp
) secs
cross apply
(
select top 1 amount
, price_date
from #sec_temp
where sec_no = secs.sec_no
order by
price_date desc
) latest
outer apply
(
select top 1 price_date
from #sec_temp
where sec_no = secs.sec_no
and amount <> latest.amount
order by
price_date desc
) previous
This prints:
sec_no price days_at_price
123ABC 25,00 5
456DEF 22,00 1