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
Related
When I execute below query SQL run this plan and it took a long time to run it and it will not be over.
QueryPlanLink
I have 3 million records in #T table.
myCode:
;WITH cte1 AS (
SELECT NationalId,len(NationalId) as LenNationalId,CustomerType,FullDateInt,time,
SUM(Price) as SUMPrice
,AVG(Price) as Price
,SUM(Volume) as Volume
,SUM (sum([Volume])) OVER (PARTITION BY NationalId,len(NationalId) ORDER BY FullDateInt,[Time]) as SumVol
,ROW_NUMBER() OVER (PARTITION BY NationalId,len(NationalId) ORDER BY FullDateInt,[Time]) AS rn
from #T as T1
group by NationalId,len(NationalId),CustomerType,FullDateInt,time
), rcte AS (
SELECT *, Price AS Cost , cast(0 as decimal) as Profit
FROM cte1 AS base
WHERE base.rn = 1
UNION ALL
SELECT curr.*, Case when curr.Volume>0 Then ((curr.Volume *curr.Price) + (prev.Cost*prev.SumVol))/nullif(curr.SumVol,0)
when curr.Volume<0 Then prev.Cost
End
as Cost
,ISNULL(Cast (Case when curr.Volume<0 Then -1*(curr.Price-Cost)*curr.Volume End as decimal),0) as Profit
FROM cte1 AS curr
INNER JOIN rcte AS prev
ON curr.NationalId = prev.NationalId AND curr.rn = prev.rn + 1
)
Select * from rcte
option (maxrecursion 0)
Is there any way to make it better?
Thanks
I Change My Query like below And Everything is Done. Thanks For All.
SELECT NationalId,len(NationalId) as LenNationalId,CustomerType,FullDateInt,time,
SUM(Price) as SUMPrice
,AVG(Price) as Price
,SUM(Volume) as Volume
,SUM (sum([Volume])) OVER (PARTITION BY NationalId,len(NationalId) ORDER BY FullDateInt,[Time]) as SumVol
,ROW_NUMBER() OVER (PARTITION BY NationalId,len(NationalId) ORDER BY FullDateInt,[Time]) AS rn
into #TCTE from #T as T1
group by NationalId,len(NationalId),CustomerType,FullDateInt,time
;With rcte AS (
SELECT *, Price AS Cost , cast(0 as decimal) as Profit
FROM #TCTE AS base
WHERE base.rn = 1
UNION ALL
SELECT curr.*, Case when curr.Volume>0 Then ((curr.Volume *curr.Price) + (prev.Cost*prev.SumVol))/nullif(curr.SumVol,0)
when curr.Volume<0 Then prev.Cost
End
as Cost
,ISNULL(Cast (Case when curr.Volume<0 Then -1*(curr.Price-Cost)*curr.Volume End as decimal),0) as Profit
FROM #TCTE AS curr
INNER JOIN rcte AS prev
ON curr.NationalId = prev.NationalId AND curr.rn = prev.rn + 1
)
Select *
into #TFinal from rcte
option (maxrecursion 0)
I have a table TRANS that contains the following records:
TRANS_ID TRANS_DT QTY
1 01-Aug-2020 5
1 01-Aug-2020 1
1 03-Aug-2020 2
2 02-Aug-2020 1
The expected output:
TRANS_ID TRANS_DT BEGBAL TOTAL END_BAL
1 01-Aug-2020 0 6 6
1 02-Aug-2020 6 0 6
1 03-Aug-2020 6 2 8
2 01-Aug-2020 0 0 0
2 02-Aug-2020 0 1 1
2 03-Aug-2020 1 0 1
Each trans_id starts with a beginning balance of 0 (01-Aug-2020). For succeeding days, the beginning balance is the ending balance of the previous day and so on.
I can create PL/SQL block to create the output. Is it possible to get the output in 1 SQL statement?
Thanks.
Try this following script using CTE-
Demo Here
WITH CTE
AS
(
SELECT DISTINCT A.TRANS_ID,B.TRANS_DT
FROM your_table A
CROSS JOIN (SELECT DISTINCT TRANS_DT FROM your_table) B
),
CTE2
AS
(
SELECT C.TRANS_ID,C.TRANS_DT,SUM(D.QTY) QTY
FROM CTE C
LEFT JOIN your_table D
ON C.TRANS_ID = D.TRANS_ID
AND C.TRANS_DT = D.TRANS_DT
GROUP BY C.TRANS_ID,C.TRANS_DT
ORDER BY C.TRANS_ID,C.TRANS_DT
)
SELECT F.TRANS_ID,F.TRANS_DT,
(
SELECT COALESCE (SUM(QTY), 0) FROM CTE2 E
WHERE E.TRANS_ID = F.TRANS_ID AND E.TRANS_DT < F.TRANS_DT
) BEGBAL,
(
SELECT COALESCE (SUM(QTY), 0) FROM CTE2 E
WHERE E.TRANS_ID = F.TRANS_ID AND E.TRANS_DT = F.TRANS_DT
) TOTAL ,
(
SELECT COALESCE (SUM(QTY), 0) FROM CTE2 E
WHERE E.TRANS_ID = F.TRANS_ID AND E.TRANS_DT <= F.TRANS_DT
) END_BAL
FROM CTE2 F
You can as well do like this (I would assume it's a bit faster): Demo
with
dt_between as (
select mindt + level - 1 as trans_dt
from (select min(trans_dt) as mindt, max(trans_dt) as maxdt from t)
connect by level <= maxdt - mindt + 1
),
dt_for_trans_id as (
select *
from dt_between, (select distinct trans_id from t)
),
qty_change as (
select distinct trans_id, trans_dt,
sum(qty) over (partition by trans_id, trans_dt) as total,
sum(qty) over (partition by trans_id order by trans_dt) as end_bal
from t
right outer join dt_for_trans_id using (trans_id, trans_dt)
)
select
trans_id,
to_char(trans_dt, 'DD-Mon-YYYY') as trans_dt,
nvl(lag(end_bal) over (partition by trans_id order by trans_dt), 0) as beg_bal,
nvl(total, 0) as total,
nvl(end_bal, 0) as end_bal
from qty_change q
order by trans_id, trans_dt
dt_between returns all the days between min(trans_dt) and max(trans_dt) in your data.
dt_for_trans_id returns all these days for each trans_id in your data.
qty_change finds difference for each day (which is TOTAL in your example) and cumulative sum over all the days (which is END_BAL in your example).
The main select takes END_BAL from previous day and calls it BEG_BAL, it also does some formatting of final output.
First of all, you need to generate dates, then you need to aggregate your values by TRANS_DT, and then left join your aggregated data to dates. The easiest way to get required sums is to use analitic window functions:
with dates(dt) as ( -- generating dates between min(TRANS_DT) and max(TRANS_DT) from TRANS
select min(trans_dt) from trans
union all
select dt+1 from dates
where dt+1<=(select max(trans_dt) from trans)
)
,trans_agg as ( -- aggregating QTY in TRANS
select TRANS_ID,TRANS_DT,sum(QTY) as QTY
from trans
group by TRANS_ID,TRANS_DT
)
select -- using left join partition by to get data on daily basis for each trans_id:
dt,
trans_id,
nvl(sum(qty) over(partition by trans_id order by dates.dt range between unbounded preceding and 1 preceding),0) as BEGBAL,
nvl(qty,0) as TOTAL,
nvl(sum(qty) over(partition by trans_id order by dates.dt),0) as END_BAL
from dates
left join trans_agg tr
partition by (trans_id)
on tr.trans_dt=dates.dt;
Full example with sample data:
alter session set nls_date_format='dd-mon-yyyy';
with trans(TRANS_ID,TRANS_DT,QTY) as (
select 1,to_date('01-Aug-2020'), 5 from dual union all
select 1,to_date('01-Aug-2020'), 1 from dual union all
select 1,to_date('03-Aug-2020'), 2 from dual union all
select 2,to_date('02-Aug-2020'), 1 from dual
)
,dates(dt) as ( -- generating dates between min(TRANS_DT) and max(TRANS_DT) from TRANS
select min(trans_dt) from trans
union all
select dt+1 from dates
where dt+1<=(select max(trans_dt) from trans)
)
,trans_agg as ( -- aggregating QTY in TRANS
select TRANS_ID,TRANS_DT,sum(QTY) as QTY
from trans
group by TRANS_ID,TRANS_DT
)
select
dt,
trans_id,
nvl(sum(qty) over(partition by trans_id order by dates.dt range between unbounded preceding and 1 preceding),0) as BEGBAL,
nvl(qty,0) as TOTAL,
nvl(sum(qty) over(partition by trans_id order by dates.dt),0) as END_BAL
from dates
left join trans_agg tr
partition by (trans_id)
on tr.trans_dt=dates.dt;
You can use a recursive query to generate the overall date range, cross join it with the list of distinct tran_id, then bring the table with a left join. The last step is aggregation and window functions:
with all_dates (trans_dt, max_dt) as (
select min(trans_dt), max(trans_dt) from trans group by trans_id
union all
select trans_dt + interval '1' day, max_dt from all_dates where trans_dt < max_dt
)
select
i.trans_id,
d.trans_dt,
coalesce(sum(sum(t.qty)) over(partition by i.trans_id order by d.trans_dt), 0) - coalesce(sum(t.qty), 0) begbal,
coalesce(sum(t.qty), 0) total,
coalesce(sum(sum(t.qty)) over(partition by i.trans_id order by d.trans_dt), 0) endbal
from all_dates d
cross join (select distinct trans_id from trans) i
left join trans t on t.trans_id = i.trans_id and t.trans_dt = d.trans_dt
group by i.trans_id, d.trans_dt
order by i.trans_id, d.trans_dt
I'm trying to find the percent change using row number with PostgreSQL but I'm running into an error where my "percent_change" column shows 0.
Here is what I have as my code.
WITH CTE AS (
SELECT date, sales, ROW_NUMBER() OVER (ORDER by date) AS rn
FROM sales_2019)
SELECT c1.date, c1.sales,
CAST(COALESCE (((c1.sales - c2.sales) * 1.0 / c2.sales) * 100, 0) AS INT) AS percent_change
FROM CTE AS c1
LEFT JOIN CTE AS c2
ON c1.date = c2.date AND c1.rn = c2.rn + 1
Here is my SQL table in case it's needed. Thank you in advance, I greatly appreciate it.
You can use LAG() for your requirement:
select
date,
sales,
round(coalesce((((sales-(lag(sales) over (order by date)))*1.0)/(lag(sales) over (order by date)))*100,0),2)
from sales_2019
or you can try with WITH clause
with cte as ( select
date,
sales,
coalesce(lag(sales) over (order by date),0) as previous_month
from sales_2019
)
select
date,
sales,
round( coalesce( (sales-previous_month)*1.0/nullif(previous_month,0),0 )*100,2)
from cte
DEMO
EDIT as per requirement in comment
with cte as ( select
date_,
sales,
ROW_NUMBER() OVER (ORDER by date_) AS rn1,
ROW_NUMBER() OVER (ORDER by date_)-1 AS rn2
from sales_2019
)
select t1.date_,
t1.sales,
round( coalesce( (t1.sales-t2.sales)*1.0/nullif(t2.sales,0),0 )*100,2)
from cte t1 left join cte t2 on t1.rn2=t2.rn1
DEMO
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
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