SQL - 找到連續兩行相同的ID

Question

我們有表是這樣的：

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 

我們想找個連續兩行，其中PERSON領域將有相同ID null值，並根據日期領域。 （如果傷者場爲null，則GROUP字段的值，反之亦然）

所以在這個例子中只有最後兩行應列出，因爲它們是同一個ID和他們之間的日期是連續

2 null GROUP2 GROUP2   06.01.2014 
2 null GROUP3 GROUP3   07.01.2014 

我正在嘗試編寫一些SQL語法，但實際上並不知道如何開始，這可能是一些複雜的表達式。我想首先要做的是根據日期獲取兩個連續的行並檢查PERSON是否爲空。

預先感謝您

Answer 1

這是使用的好地方lag()和lead()：

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); 

編輯：

以上不相當的工作，因爲NULL將是返回每個id的第一行或最後一行。糟糕！這裏是一個修復：

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) 
    ); 

編輯II;

如何尋找兩個非空的組？

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是一個非常糟糕的名稱，因爲它是一個SQL保留字。