Question

我需要从交易表中获取交易表的交易，这些交易完成了。并检查，如果最后一天的交易金额超过前一天交易金额的10％。我的表有列AccountId，SubAccountId，Amount，Date和UserId。例如：

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),
;

在此示例中，我只需选择日期06/15/2018和06/14/2018的交易，并显示当天的交易金额总和。到目前为止，我可以选择最后的交易，如下：

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

这将通过一行上的accountID获取每天的交易和前一天的交易。从这里你可以比较价值。

Answer 2

这会检查当天的sum amount与前一天的sum amount（确认其大于10％）然后执行top 2提取仅过去两天......

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

你想按日期分组并总结金额

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

Answer 4

你可以用它。我希望它适合你。

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

您可以查看以下查询，以获取这两个日期的最后两个日期和金额总和。

选择明显的accountid，subaccountid，user，trandate，sum（amount）over（按日期划分）来自交易其中date＆gt; =（从日期＆lt;（从交易中选择max（date））的交易中选择max（date））;

按日期从Transaction table获取最后的交易

5 个答案: