Question

我们有这样的表：

ID  PERSON  GROUP  ASSIGNEDGROUP  CHANGEDATE
1   null    GROUP1 GROUP1         01.01.2014
1   NAME1   null   GROUP1         02.01.2014
1   null    GROUP2 GROUP2         03.01.2014
2   null    GROUP1 GROUP1         04.01.2014
2   NAME1   null   GROUP1         05.01.2014
2   null    GROUP2 GROUP2         06.01.2014
2   null    GROUP3 GROUP3         07.01.2014

我们希望找到两个连续的行，其中PERSON字段对于相同的ID并且基于日期字段将具有值null。（如果PERSON字段为null，则GROUP字段具有值，反之亦然）

因此，对于此示例，只应列出最后两行，因为它们用于相同的ID，并且它们之间的日期是连续的

2   null    GROUP2 GROUP2         06.01.2014
2   null    GROUP3 GROUP3         07.01.2014

我正在尝试编写一些SQL语法，但实际上不知道如何启动，这可能是一些复杂的表达式。我想首先要做的是根据日期获得两个连续的行，然后检查PERSON是否为空。

提前谢谢

Answer 1

这是一个使用lag()和lead()的好地方：

select t.*
from (select t.*,
             lag(person) over (partition by id order by changedate) as person_prev,
             lead(person) over (partition by id order by changedate) as person_next
      from table t
     ) t
where person is null and
      (person_prev is null or person_next is null);

编辑：

上述内容并不常用，因为每个NULL的第一行或最后一行都会返回id。糟糕！这是一个修复：

select t.*
from (select t.*,
             lag(person) over (partition by id order by changedate) as person_prev,
             lead(person) over (partition by id order by changedate) as person_next,
             lag(id) over (partition by id order by changedate) as id_prev,
             lead(id) over (partition by id order by changedate) as id_next
      from table t
     ) t
where person is null and
      ((person_prev is null and id_prev is not null) or
       (person_next is null and id_next is not null)
      );

编辑II;

如何寻找两个非空的组？

select t.*
from (select t.*,
             lag(group) over (partition by id order by changedate) as group_prev,
             lead(group) over (partition by id order by changedate) as group_next
      from table t
     ) t
where group is not null and
      (group_prev is not null or group_next is not null);

注意：group是列的一个非常糟糕的名称，因为它是一个SQL保留字。

SQL - 为同一个ID

1 个答案: