Question

我的结果集如下所示：

ID  Pass    DESCRIPTION DATE    STATUS
123 3455    RUNNING 13/09/2017  INSERTED
123 3455    RUNNING 14/09/2017  INSERTED
123 3455    RUNNING 15/09/2017  INSERTED
123 3455    RUNNING 16/09/2017  INSERTED
123 3455    RUNNING 17/09/2017  DELETED
123 3455    JUMPING 17/09/2017  INSERTED
123 3455    JUMPING 17/09/2017  DELETED
456 8968    JUMPING 18/09/2017  INSERTED
456 8968    JUMPING 18/09/2017  DELETED
456 8968    JUMPING 06/10/2017  INSERTED
456 8968    JUMPING 07/10/2017  INSERTED
456 8968    JUMPING 08/10/2017  INSERTED
456 8968    JUMPING 09/10/2017  RETIRADO

我希望从这个结果集中获得：

ID  Pass    DESCRIPTION DATE    STATUS
123 3455    RUNNING 13/09/2017  INSERTED
123 3455    RUNNING 17/09/2017  DELETED
123 3455    JUMPING 17/09/2017  INSERTED
123 3455    JUMPING 17/09/2017  DELETED
456 8968    JUMPING 18/09/2017  INSERTED
456 8968    JUMPING 18/09/2017  DELETED
456 8968    JUMPING 06/10/2017  INSERTED
456 8968    JUMPING 09/10/2017  RETIRADO

这些事件的最小和最大日期，忽略最小值和最大值内的所有日期。 ID，Pass和DESCRIPTION可以相同，但是当STATUS发生变化时，我希望得到最小和最大结果。有人能帮助我吗？

Answer 1

尝试使用以下

;WITH CTE
AS
(
    SELECT
        RN = ROW_NUMBER() OVER(PARTITION BY ID,Pass,Description,Status ORDER BY Date ASC),
        *
        FROM T1
)
SELECT
  *
  FROM CTE
    WHERE RN = 1

检查Demo此处

Answer 2

您可以使用lag获取相关结果集。

select * 
from (select t.*,lag(status) over(partition by id order by date_column) as prev_status
      from tbl t
     ) t
where prev_status is null or prev_status <> status

编辑：由于给定日期有多行，您应指定一列以便可靠地排序行。使用lag(status) over(partition by id order by date_column,other_order_column)获得一致的结果。

试图从查询中获取特定结果

2 个答案: