SQL 服务器：在 table 中查找两列不连续的行

Question

我有一个 table 数据如下所示：

pK    Customer     DateTime1                  DateTime2
1     6            2016-04-01 00:00:00.000    2016-10-09 00:00:00.000
2     6            2016-07-01 00:00:00.000    2016-10-21 00:00:00.000
3     6            2016-10-01 00:00:00.000    2016-10-20 00:00:00.000

我想查找在订购 DateTime1 时相应的 DateTime2 值（按客户 ID 筛选时）不遵循相同顺序的行。

所以在上面的例子中，我想找到 pK 为 3 的行，因为当 DateTime1 按升序排序时，DateTime2 不大于第 2 行中的 DateTime2。

看起来与这个问题类似，但它处理的是项目的顺序而不是不等式： TSQL check if specific rows sequence exists

我尝试使用 CTE 语句的一个版本

Answer 1

这似乎是 row_number() 的一个很好的应用：

select t.*
from (select t.*,
             row_number() over (partition by customer order by datetime1) as seqnum_1,
             row_number() over (partition by customer order by datetime2) as seqnum_2
      from t
     ) t
where seqnum_1 <> seqnum_2;

但这会 return 所有排序错误的行，基于全局排序（在本例中为 pk 2 和 3）。

您只需要在给定行上改变方向的位置。为此，使用 lag():

select t.*
from (select t.*,
             lag(datetime1) over (partition by customer order by pk) as prev_dt1,
             lag(datetime2) over (partition by customer order by pk) as prev_dt2
      from t
     ) t
where (dt1 > prev_dt1 and dt2 <= prev_dt2) or
      (dt1 < prev_dt1 and dt2 >= prev_dt2);

Answer 2

Declare @YourTable table (pK int,Customer int,DateTime1 datetime,DateTime2 datetime)
Insert Into @YourTable values
(1,6,'2016-04-01 00:00:00.000','2016-10-09 00:00:00.000'),
(2,6,'2016-07-01 00:00:00.000','2016-10-21 00:00:00.000'),
(3,6,'2016-10-01 00:00:00.000','2016-10-20 00:00:00.000')

;with cte as (
    Select *,Flg=Row_Number() over (Partition By Customer Order By DateTime1) - Row_Number() over (Partition By Customer Order By DateTime2) 
     From  @YourTable
)
Select pK
      ,Customer
      ,DateTime1
      ,DateTime2 
 From  cte 
 Where Flg>0

Returns

pK  Customer    DateTime1               DateTime2
3   6           2016-10-01 00:00:00.000 2016-10-20 00:00:00.000

SQL 服务器：在 table 中查找两列不连续的行

SQL Server: Find rows in table where both columns not sequential

sql

sql-server

sequential