I need to get Transactions from Transaction Table from 2 lats dates, where this Transactions completed. And check, if amount of transaction on last day more than 10% than amount of transaction for previous day.
My table Have columns AccountId, SubAccountId, Amount, Date and UserId.
For example:
CREATE TABLE Transactions
(`id` int, `AccountId` int, `SubAccountId` int, `Amount` decimal
,`Date` datetime, `User` int);
INSERT INTO Transactions
(`id`, `AccountId`, `SubAccountId`, `Amount`, `Date`, `User`)
VALUES
(1, 1, 2, 100, '06/15/2018', 1),
(2, 1, 2, 40, '06/15/2018', 1),
(3, 1, 2, 20, '06/14/2018', 1),
(4, 1, 2, 0, '06/10/2018', 1),
;
In this example I need to select only transactions for date 06/15/2018 and 06/14/2018, and display sum of amount of transactions for this days.
So far, I can select the last transactions, like this:
select distinct AccountId,
SubAccountId,
UserId,
Amount,
Date AS lastDate,
min(Date)
over (partition by PayerAccount order by Date
rows between 1 preceding and 1 preceding) as PrevDate
from Transactions
order by UserId
This checks the sum amount of the current day against the sum amount of the previous day (to confirm it's greater than 10%) and then does a top 2 to extract only the last two days...
WITH CTE AS(
select
Date,
sum(Amount) as SumAmount,
rownum = ROW_NUMBER() OVER(ORDER BY Date)
from Transactions
group by Date
)
select top 2 CTE.date, CTE.SumAmount, CTE.rownum, CASE WHEN prev.sumamount > CTE.sumamount * 0.10 THEN 1 else 0 END isgreaterthan10per
from CTE
LEFT JOIN CTE prev ON prev.rownum = CTE.rownum - 1
order by CTE.date desc
with CTE1 as
(
select accountID, Date, sum(Amount) as Amount
from Transactions
where Date between '2018-06-14' and '2018-06-16' -- Apply date restriction here
group by accountID, Date
)
, CTE2 as
(
select accountID, Amount, Date,
row_number() over (partition by accountID order by date desc) as rn
from Transactions
)
select a1.accountID, a1.Amount, a1.Date, a2.Date, a2.Amount
from CTE2 a1
left join CTE2 a2
on a1.accountID = a2.accountID
and a2.rn = a1.rn+1
This will get you the transactions for each day and those for the day previous by accountID on one line. From here you can compare values.
you wanna group by date and sum the amount
select Date,sum(Amount) from Transactions /*where contitions*/ group by Date
You can use this. I hope it will for you.
SELECT
*
FROM Transactions tb
INNER JOIN
(
SELECT MAX([Date]) AS [Date] FROM Transactions
UNION ALL
SELECT MAX([Date]) AS [Date] FROM Transactions WHERE [Date] < (SELECT MAX([Date]) AS [Date] FROM Transactions)
) tb1 ON tb1.[Date] = tb.[Date]
You can check out below query to get last two date and sum of amount on those two dates.
select distinct accountid,subaccountid,user,trandate,sum(amount) over(partition by date)
from transactions
where date>=(select max(date) from transactions where date < (select max(date) from transactions));
Related
I have a DATE, ORDERID, CLIENTID and STOREID.
I'm trying to get the average number of days between the FIRST order date and THIRD order date for all clients who have placed at least 3 orders.
This is what I have so far but when I add OrderId, it doesn’t return anything anymore.
select Date, OrderId, ClientId
from ClientOrders
group by Date, OrderId, ClientId
having count(ClientId) > 3
We can use ROW_NUMBER here:
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY ClientId ORDER BY Date) rn
FROM ClientOrders
)
SELECT AVG(diff)
FROM
(
SELECT DATEDIFF(day,
MAX(CASE WHEN rn = 1 THEN Date END),
MAX(CASE WHEN rn = 3 THEN Date END)) AS diff
FROM cte
GROUP BY ClientId
HAVING COUNT(*) >= 3 -- at least 3 orders
) t;
I trying to take an average per customer, but you're not grouping by customer.
I would like to calculate the average days between several order dates from a table called invoice. For each BusinessPartnerID, what is the average days between orders i want average days last three records orders .
I got the average of all order for each user but need days last three records orders
The sample table is as below
;WITH temp (avg,invoiceid,carname,carid,fullname,mobail)
AS
(
SELECT AvgLag = AVG(Lag) , Lagged.idinvoice,
Lagged.carname ,
Lagged.carid ,Lagged.fullname,Lagged.mobail
FROM
(
SELECT
(car2.Name) as carname ,
(car2.id) as carid ,( busin.Name) as fullname, ( busin.Mobile) as mobail , INV.Id as idinvoice , Lag = CONVERT(int, DATEDIFF(DAY, LAG(Date,1)
OVER (PARTITION BY car2.Id ORDER BY Date ), Date))
FROM [dbo].[Invoice] AS INV
JOIN [dbo].[InvoiceItem] AS INITEM on INV.Id=INITEM.Invoiceid
JOIN [dbo].[BusinessPartner] as busin on busin.Id=INV.BuyerId and Type=5
JOIN [dbo].[Product] as pt on pt.Id=INITEM.ProductId and INITEM.ProductId is not null and pt.ProductTypeId=3
JOIN [dbo].[Car] as car2 on car2.id=INv.BusinessPartnerCarId
) AS Lagged
GROUP BY
Lagged.carname,
Lagged.carid,Lagged.fullname,Lagged.mobail, Lagged.idinvoice
-- order by Lagged.fullname
)
SELECT * FROM temp where avg is not null order by avg
I don't really see how your query relate to your question. Starting from a table called invoice that has columns businesspartnerid, and date, here is how you would take the average of the day difference between the last 3 invoices of each business partner:
select businesspartnerid,
avg(1.0 * datediff(
day,
lag(date) over(partition by businesspartnerid order by date),
date
) avg_diff_day
from (
select i.*,
row_number() over(partiton by businesspartnerid order by date desc) rn
from invoice i
) i
where rn <= 3
group by businesspartnerid
Note that 3 rows gives you 2 intervals only, that will be averaged.
I have below table called orders which has customer id and their order date (Note: there can be multiple orders from same customer on a single day)
create table orders (Id char, order_dt date)
insert into orders values
('A','1/1/2020'),
('B','1/1/2020'),
('C','1/1/2020'),
('D','1/1/2020'),
('A','1/1/2020'),
('B','1/1/2020'),
('A','2/1/2020'),
('B','2/1/2020'),
('C','2/1/2020'),
('B','2/1/2020'),
('A','3/1/2020'),
('B','3/1/2020')
I'm trying to write an SQL query to find the number of customers who shopped for 3 consecutive days in month of January 2020
Based on above order values, the output should be: 2
I referred other similar questions but still wasn't able to come the exact solution
Here is my solution which works fine even there are many orders of one customer in one day;
Some scripts to build test environment:
create table orders (Id varchar2(1), order_dt date);
insert into orders values('A',to_date('01/01/2020','dd/mm/yyyy'));
insert into orders values('B',to_date('01/01/2020','dd/mm/yyyy'));
insert into orders values('C',to_date('01/01/2020','dd/mm/yyyy'));
insert into orders values('D',to_date('01/01/2020','dd/mm/yyyy'));
insert into orders values('A',to_date('01/01/2020','dd/mm/yyyy'));
insert into orders values('B',to_date('01/01/2020','dd/mm/yyyy'));
insert into orders values('A',to_date('02/01/2020','dd/mm/yyyy'));
insert into orders values('B',to_date('02/01/2020','dd/mm/yyyy'));
insert into orders values('C',to_date('02/01/2020','dd/mm/yyyy'));
insert into orders values('B',to_date('02/01/2020','dd/mm/yyyy'));
insert into orders values('A',to_date('03/01/2020','dd/mm/yyyy'));
insert into orders values('B',to_date('03/01/2020','dd/mm/yyyy'));
select distinct id, count_days from (
select id,
order_dt,
count(*) over(partition by id order by order_dt range between 1 preceding and 1 following ) count_days
from orders group by id, order_dt
)
where count_days = 3;
-- Insert for test more days than 3 consecutive
insert into orders values('A',to_date('04/01/2020','dd/mm/yyyy'));
You can use two window functions to calculate difference between consequtive dates and sliding window with ROWS offset to count distinct preceiding consequtive days. Example here:
with gen as (
select 1 as cust_id, (date '2020-01-10') + 1 as q from dual union all
select 1, (date '2020-01-10') + 2 as q from dual union all
select 1, (date '2020-01-10') + 3 as q from dual union all
select 1, (date '2020-01-10') + 3 as q from dual union all
select 1, (date '2020-01-10') + 5 as q from dual union all
select 1, (date '2020-01-10') + 7 as q from dual union all
select 1, (date '2020-01-10') + 8 as q from dual union all
select 1, (date '2020-01-10') + 9 as q from dual
)
, diff as (
select gen.*
, q - lag(q) over(partition by cust_id, trunc(q, 'mm') order by q asc) as datediff
from gen
)
, window as (
select diff.*
, sum(decode(datediff, 1, 1, 0)) over(partition by cust_id, trunc(q, 'mm') order by q asc range between 2 preceding and current row) as cnt
from diff
)
select sum(count(distinct q)) as cnt
from window
where cnt = 2
group by cust_id
why not join twice based on same following two days. As long as you have index on the customer's ID and date, the join should be optimized. Because the joins require match on the same starting date basis, it either finds or it doesn't. If not, it is left out of the result set.
select distinct
o1.id
from
orders o1
JOIN orders o2
on o1.id = o2.id
AND o1.order_dt = o2.order_dt - interval '1' day
JOIN orders o3
on o1.id = o3.id
AND o1.order_dt = o3.order_dt - interval '2' day
Hmmmm . . . one method is to use lead()/lag(). Assuming that you don't have duplicates on a single day, then:
select distinct id
from (select o.*,
lag(order_dt) over (partition by id order by order_dt) as prev_order_dt,
lag(order_dt, 2) over (partition by id order by order_dt) as prev_order_dt2
from orders o
where order_dt >= date '2020-01-01' and
order_dt < date '2020-02-01'
) o
where prev_order_dt = order_dt - interval '1' day and
prev_order_dt2 = order_dt - interval '2' day;
EDIT:
If the table has duplicate records, the above is easily tweaked:
select distinct id
from (select o.*,
lag(order_dt) over (partition by id order by order_dt) as prev_order_dt,
lag(order_dt, 2) over (partition by id order by order_dt) as prev_order_dt2
from (select distinct o.id, trunc(order_dt) as order_dt
from orders o
where order_dt >= date '2020-01-01' and
order_dt < date '2020-02-01'
) o
) o
where prev_order_dt = order_dt - interval '1' day and
prev_order_dt2 = order_dt - interval '2' day;
I have to query in sql server where I have to find for each id it's volume such that we have last 1 year date for each id with it's volume.
for example below is my data ,
for each id I need to query the last 1 year transaction from when we have the entry for that id as you can see from the snippet for id 1 we have the latest date as 7/31/2020 so I need the last 1 year entry from that date for that id, The highlighted one is exclude because that date is more than 1 year from the latest date for that id
Similarly for Id 3 we have all the date range in one year from the latest date for that particular id
I tried using the below query and I can get the latest date for each id but I am not sure how to extract all the dates for each id from the latest date to one year, I would appreciate if some one could help me.
I am using Microsoft sql server would need the query which executes in sql server, Table name is emp and have millions of id
Select *
From emp as t
inner join (
Select tm.id, max(tm.date_tran) as MaxDate
From emp tm
Group by tm.id
) tm on t.id = tm.id and t.date_tran = tm.MaxDate
To exclude transactions where the date difference between the tran_date and the maximum tran_date for each id is greater than 1 year, something like this:
;with max_cte(id, max_date) as (
Select id, max(date_tran)
From emp tm
Group by id )
Select *
From emp e
join max_cte mc on e.id=mc.id
and datediff(d, e.date_tran, mc.max_date)<=365;
Update: per comments, added volume. Thnx GMB :)
;with max_cte(id, date_tran, volume, max_date) as (
Select *, dateadd(year, -1, max(date_tran) over(partition by id)) max_date
From #emp tm)
Select id, sum(volume) sum_volume
From max_cte mc
where mc.date_tran>max_date
group by id;
You can do this with window functions:
select id, sum(volume) total_volume
from (
select t.*, max(date_tran) over(partition by id) max_date_tran
from mytable t
) t
where date_tran > dateadd(year, -1, max_date_tran)
group by id
Alternatively, you can use a correlated subquery for filtering:
select id, sum(volume) total_volume
from mytable t
where t.date_tran > (
select dateadd(year, -1, max(t1.date_tran))
from mytable t1
where t1.id = t.id
)
The second query would take advantage of an index on (id, date_tran).
this should do the trick for you:
SELECT
*
FROM
emp
JOIN
(
SELECT
MAX(date_tran) max_date_tran
, Id
FROM
emp
GROUP BY
id
) emp2
ON emp2.Id = emp.Id
AND DATEADD(YEAR, -1, emp2.max_date_tran) <= emp.date_tran;
Your code is good. Just add the date difference function to get the particular time in between the transaction, like the following:
Select *
From emp as t
inner join ( Select id as id, max(date_tran) as maxdate
From emp tm
Group by id
) tm on t.id = tm.id and datediff(d, e.date_tran, mc.maxdate)<=365;
I have table with fields Customer.No. , Posting date, Order_ID . I want to find total inactive customers for last 12 months on month basis which means they have placed order before 12 months back and became in active. So want calculate this every month basis to under stand how inactive customers are growing month by month.
if I run the query in July it should go back 365 days from the previous month end and give total number of inactive customers. I want to do this month by month.
I am in learning stage please help.
Thanks for your time in advance.
to get the customers
SELECT DISTINCT a.CustomerNo
FROM YourTable a
WHERE NOT EXISTS
(SELECT 0 FROM YourTable b WHere a.CustomerNo = b.CustomerNo
and b.PostingDate >
dateadd(day,-365 -datepart(day,getdate()),getdate())
)
to get a count
SELECT DISTINCT count(0) as InnactiveCount
FROM YourTable a
WHERE NOT EXISTS
(SELECT 0 FROM YourTable b WHere a.CustomerNo = b.CustomerNo
and b.PostingDate >
dateadd(day,-365 -datepart(day,getdate()),getdate())
..
generate a 'months' table by CTE, then look for inactive in those months
;WITH month_gen as (SELECT dateadd(day,-0 -datepart(day,getdate()),getdate()) eom, 1 as x
UNION ALL
SELECT dateadd(day,-datepart(day,eom),eom) eom, x + 1 x FROM month_gen where x < 12
)
SELECT DISTINCT CONVERT(varchar(7), month_gen.eom, 102), count(0) innactiveCount FROM YourTable a
cross join month_gen
WHERE NOT EXISTS(SELECT 0 FROM YourTable b WHere a.CustomerNo = b.CustomerNo and
YEAR(b.PostingDate) = YEAR(eom) and
MONTH(b.PostingDate) = MONTH(eom)
)
GROUP BY CONVERT(varchar(7), month_gen.eom, 102)
if that gets you anywhere, maybe a final step is to filter out anything getting 'counted' before it was ever active i.e. don't count 'new' customers before they became active
Try below query. To achieve your goal you need calendar table (which I defined with CTE). Below query counts inactivity for the first day of a month:
declare #tbl table (custNumber int, postDate date, orderId int);
insert into #tbl values
(1, '2017-01-01', 123),
(2, '2017-02-01', 124),
(3, '2017-02-01', 125),
(1, '2018-02-02', 126),
(2, '2018-05-01', 127),
(3, '2018-06-01', 128)
;with cte as (
select cast('2018-01-01' as date) dt
union all
select dateadd(month, 1, dt) from cte
where dt < '2018-12-01'
)
select dt, sum(case when t2.custNumber is null then 1 else 0 end)
from cte c
left join #tbl t1 on dateadd(year, -1, c.dt) >= t1.postDate
left join #tbl t2 on t2.postDate > dateadd(year, -1, c.dt) and t2.postDate <= c.dt and t1.custNumber = t2.custNumber
group by dt