按日期从交易表中获取最后一笔交易

Question

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

Answer 1

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.

Answer 2

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

Answer 3

您想按日期分组并求和

select Date,sum(Amount) from Transactions /*where contitions*/ group by Date

Answer 4

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]

Answer 5

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));