Question

SQL数据库。

我必须确定发生的重复交易清单，范围为5-6分钟。

BillId  Trans   DateTime

A100125  Paid  2018-04-18 11:21:40.873 - Valid Transaction
A100125  Paid  2018-04-18 11:24:40.873 - Duplicate Transaction
A100125  Paid  2018-04-18 11:30:40.873 - Duplicate Transaction
A100125  Paid  2018-04-18 12:30:40.873 - Valid Transaction

我可以为日期范围生成报告。这大约是5-10分钟的平均差异。

Answer 1

这是

简而言之，case语句中存在一个比较每一行的子查询。这不是最有效的，因为约瑟夫不是sargable

declare @tabel Table (BillId varchar(8000),  Trans varchar(50), DateOfSomethingProbablyTheTransaction   DateTime)
declare @interval int = 6
insert into @tabel
values ('A100125',  'Paid',  '2018-04-18 11:21:40.873'), --valid
('A100125',  'Paid',  '2018-04-18 11:24:40.873'), -- Duplicate Transaction
('A100125',  'Paid',  '2018-04-18 11:30:40.873'), -- Duplicate Transaction
('A100125',  'Paid',  '2018-04-18 12:30:40.873') -- Valid Transaction)

select *, 
case when exists(select 1 
                from @tabel t2 
                where t1.BillId = t2.BillId 
                and t1.Trans = t2.Trans 
                and DATEDIFF(MINUTE, t2.DateOfSomethingProbablyTheTransaction, t1.DateOfSomethingProbablyTheTransaction) <= @interval
                and t1.DateOfSomethingProbablyTheTransaction > t2.DateOfSomethingProbablyTheTransaction)
 then 'Invalid' else'valid' end as Validity
 from @tabel t1

另请注意，如果您希望它看起来更好，因为查询优化器只是在场景后面进行连接，如果trans状态无关紧要，可以将其删除以检查重复项 p>

Answer 2

我偷了其他人的设置：

但这是一个使用滞后的答案：

declare @tabel Table (BillId varchar(8000),  Trans varchar(50), DateOfSomethingProbablyTheTransaction   DateTime)
declare @interval int = 6
insert into @tabel
values ('A100125',  'Paid',  '2018-04-18 11:21:40.873'), --valid
('A100125',  'Paid',  '2018-04-18 11:24:40.873'), -- Duplicate Transaction
('A100125',  'Paid',  '2018-04-18 11:30:40.873'), -- Duplicate Transaction
('A100125',  'Paid',  '2018-04-18 12:30:40.873') -- Valid Transaction)

(
select BillId,  Trans,   DateOfSomethingProbablyTheTransaction dt
    ,lag(DateOfSomethingProbablyTheTransaction,1) over (partition by BillID,Trans order by DateOfSomethingProbablyTheTransaction)
    ,case when datediff(minute
                       ,lag(DateOfSomethingProbablyTheTransaction,1) over (partition by BillID,Trans order by DateOfSomethingProbablyTheTransaction)
                       ,DateOfSomethingProbablyTheTransaction)<=6
        then 'Invalid'
        else 'Valid'
     end DataCheck
from @tabel
)

结果：

BillId  Trans   dt                      (No column name)        DataCheck
A100125 Paid    2018-04-18 11:21:40.873 NULL                    Valid
A100125 Paid    2018-04-18 11:24:40.873 2018-04-18 11:21:40.873 Invalid
A100125 Paid    2018-04-18 11:30:40.873 2018-04-18 11:24:40.873 Invalid
A100125 Paid    2018-04-18 12:30:40.873 2018-04-18 11:30:40.873 Valid

再次使用CTE :(不确定何时开始LAG）

;with base as
(
select BillId,  Trans,   DateOfSomethingProbablyTheTransaction dt
    ,rn = row_number() over (partition by BillId,  Trans order by DateOfSomethingProbablyTheTransaction)
from @tabel
)

select base.*
        ,prior.dt
    ,Test = case when datediff(minute,isnull(prior.dt,'1/1/1900'),base.dt) <=6 then 'Invalid' else 'Valid' end
from base
    left join base as [prior] on base.rn-1 = [prior].rn
            and base.BillId=[prior].BillId
            and base.Trans = [prior].trans

Answer 3

我不理解你需要什么，但这可能会有所帮助。这使得同一账单id上的交易具有不同的时间，其中时间差小于7分钟。

select *
from Transactions T1
INNER JOIN Transactions T2
ON T1.BillId=T2.BillId
AND T1.DateTime<>T2.DateTime
AND DATEDIFF(MI,T1.DateTime,T2.DateTime)<7

如果我误解了，请告诉我。

Answer 4

毫无疑问，有更好的方法可以做到这一点。但是，无论如何，我们走了。

SELECT q1.Billid, q1.Trans, q1.TranTime , 
       CONVERT(INT, Isnull(Datediff(minute, q2.TranTime, q1.TranTime), 0)) AS [Difference], 
       CASE 
         WHEN CONVERT(INT, Isnull(Datediff(minute, q2.TranTime, q1.TranTime), 100)) > 6 THEN 'Valid'  -- (diff > 6) || (diff = 0)
         ELSE 'Invalid' 
       END as Validity
FROM   (SELECT *,   
       Row_number() OVER (ORDER BY [Billid], [Trans], TranTime DESC) AS rn 
       FROM   tablename2) q1 
       LEFT JOIN ((SELECT *, 
                   Row_number()  OVER (ORDER BY [Billid], [Trans], TranTime DESC) AS rn 
                   FROM   tablename2)) q2 
       ON q1.rn = q2.rn - 1 
ORDER  BY TranTime ASC

结果

+---------+-------+-------------------------+------------+----------+
| Billid  | Trans |        TranTime         | Difference | validity |
+---------+-------+-------------------------+------------+----------+
| A100125 | Paid  | 2018-04-18 11:21:40.873 |          0 | Valid    |
| A100125 | Paid  | 2018-04-18 11:24:40.873 |          3 | Invalid  |
| A100125 | Paid  | 2018-04-18 11:30:40.873 |          6 | Invalid  |
| A100125 | Paid  | 2018-04-18 12:30:40.873 |         60 | Valid    |
| A100125 | Paid  | 2018-04-18 12:31:40.873 |          1 | Invalid  |
+---------+-------+-------------------------+------------+----------+

如何在5-10分钟的时差内获取重复记录？

4 个答案: