Question

partition key | source id | row start date | row end date | row is current | id
--------------+-----------+----------------+--------------+----------------+-------
20150130001   | 1         | 1900-01-01     | 2015-01-30   | 0              | 1
20150131001   | 1         | 2015-01-31     | 2015-01-31   | 0              | 1
20150201001   | 1         | 2015-02-01     | 2015-02-01   | 0              | 1
20150213001   | 1         | 2015-02-02     | 2015-02-11   | 0              | 1
20150223001   | 1         | 2015-02-14     | 2015-02-23   | 0              | 1
20150303001   | 1         | 2015-02-24     | 2015-02-25   | 0              | 1
99991231001   | 1         | 2015-03-04     | 9999-12-31   | 1              | 1

如何找到行结束日期和下一行开始日期的行与同一个ID的差距> 1？

Answer 1

使用LEAD或LAG功能。您可以找到更多示例here

将Table名称从Table1更改为您的表名，您应该得到一些结果。

With tblDates as (
select *,
        LEAD( [row start date], 1) OVER(Order by [row start date]) as NextStartDate
from Table1 )

select * 
from tblDates
Where DATEDIFF(d, [row end date], NextStartDate) > 1

Answer 2

有了这类问题，我立即想到了EXISTS（）。假设您的日期字段是datetime数据类型：

SELECT *
FROM MyTable t1
WHERE NOT EXISTS(
  SELECT * FROM MyTable t2
  WHERE t1.id=t2.id
  AND DATEDIFF(dd, t1.[Row End Date], t2.[Row Start Date]) BETWEEN 0 AND 1
)

Answer 3

如果您不使用SQL Server 2012+，则可以使用CROSS APPLY：

select *
from Table1 A
cross apply (
     select min(rowstartdate) 
     from Table1 where rowstartdate > A.rowenddate
     where id = A.id
) B(Minimo)
where DateDiff(DAY, A.rowenddate, B.Minimo) > 1

SQL Fiddle

Answer 4

我发现最简单的方法是：

SELECT *
FROM yourTable t
WHERE 
    DATEDIFF(DAY, t.endDate, (SELECT MIN(ti.startDate) FROM yourTable ti WHERE ti.partitionKey > t.partitionKey)) > 1

此查询只是说：当天行结束日期与当天下一个分区键的最低（第一个）开始日期之间存在差异的所有行都大于一（日）。

如何查找行结束日期和下一行开始日期的行对于相同的ID具有间隙＆gt; 1

4 个答案: