Question

我有一个具有重复值的表。我想做的是汇总各列。我唯一的问题是我不想基于表中的列NewTransaction对所有要重复的重复值求和。

RecoveryKey   DateTime      Duration NewTransaction Variable8   CallTypeID
--------------------------------------------------------------------------
7994113912466 12/4/18 16:26 19       Y              152643-5657     -1    
7994113912470 12/4/18 16:26 1168     Y              152643-5657   6390
7994113912751 12/4/18 16:51 2686     N              152643-5657   6390

所以从表中可以看到，我要为Max(DateTime)选择Variable8的记录是具有NewTransaction == Y的重复记录。

与具有NewTransaction == N

的记录相似

如果您这样做，得到的值将低于-

RecoveryKey   DateTime      Duration NewTransaction Variable8   CallTypeID
--------------------------------------------------------------------------
7994113912470 12/4/18 16:26 1168     Y              152643-5657    6390
7994113912751 12/4/18 16:51 2686     N              152643-5657    6390

为此，我需要的是max(RecoveryKey)，max(DateTime)，Sum(Duration)。我的最终结果将如下所示

RecoveryKey   DateTime      Duration NewTransaction Variable8   CallTypeID
---------------------------------------------------------------------------
7994113912470 12/4/18 16:26 3854     Y              152643-5657    6390

Answer 1

我想这就是你想要的

select max(RecoveryKey)
  , distinct DateTime
  , Sum(Duration)
  , NewTransaction, Variable8
  , CallTypeID
from MyTable
group by NewTransaction, Variable8
having datetime = max(datetime);

Answer 2

基于这个问题，我相信您正在尝试做以下两件事：

针对每种max([DateTime])和Variable8组合，向下过滤一条记录，即NewTransaction。
让各列显示NewTransaction值为Y的记录中包含的数据，而不显示通过步骤1的筛选器得到的记录中的总Duration 。

答案

with initial_filter as
    (
        select *
        , row_number() over (partition by Variable8, NewTransaction order by [DateTime] desc, RecoveryKey desc) as rnk_a
        , row_number() over (partition by Variable8 order by NewTransaction desc, [DateTime] desc, RecoveryKey desc) as rnk_b
        from MyTable as t
    )
select max(b.RecoveryKey) as RecoveryKey
, max(b.[DateTime]) as [DateTime]
, sum(a.Duration) as Duration
, max(b.NewTransaction) as NewTransaction
, max(b.Variable8) as Variable8
, max(b.CallTypeID) as CallTypeID
from initial_filter as a
left join initial_filter as b on a.RecoveryKey = b.RecoveryKey and b.rnk_b = 1
where a.rnk_a = 1

结果：

+---------------+-------------------------+----------+----------------+-------------+------------+
| RecoveryKey   | DateTime                | Duration | NewTransaction | Variable8   | CallTypeID |
+---------------+-------------------------+----------+----------------+-------------+------------+
| 7994113912470 | 2018-12-04 16:26:00.000 | 3854     | Y              | 152643-5657 | 6390       |
+---------------+-------------------------+----------+----------------+-------------+------------+

Answer 3

您好Abdulquadir Shaikh，

也许这就是您想要的。

create table test22
(
RecoveryKey varchar(64),
[DateTime] datetime,
Duration int,
NewTransaction varchar(30),
Variable8 varchar(64),
CallTypeID int
)

insert into test22 values
('7994113912466','12/4/18 16:26:00',19,'Y','152643-5657',-1),
('7994113912470','12/4/18 16:26:00',1168,'Y','152643-5657',6390),
('7994113912751','12/4/18 16:51:00',2686,'N','152643-5657',6390)

;WITH CTE AS
(
select
max(RecoveryKey) as RecoveryKey,
max([DateTime]) as [DateTime],
max(Duration) as Duration,
NewTransaction,
Variable8,
max(CallTypeID) as CallTypeID
from test22
group by Variable8,NewTransaction
)
SELECT 
MAX(RecoveryKey) AS RecoveryKey,
MAX([DateTime]) AS [DateTime],
SUM(Duration) AS Duration,
MAX(NewTransaction) AS NewTransaction,
Variable8,
MAX(CallTypeID) AS CallTypeID
FROM CTE
GROUP BY Variable8

--Output
/*
RecoveryKey                                                      DateTime                Duration    NewTransaction                 Variable8                                                        CallTypeID
---------------------------------------------------------------- ----------------------- ----------- ------------------------------ ---------------------------------------------------------------- -----------
7994113912751                                                    2018-12-04 16:51:00.000 3854        Y                              152643-5657                                                      6390
*/

如果对我的答复有任何疑问，请随时告诉我。

最好的问候，

会

根据SQL Server中的条件进行组重复

3 个答案: