oracle sql使用不同行类型的单个日期列查找最小值和最大值

Question

INPUT：

+--------+---------+------+------+------+
| EMP_ID |  DT_1   | ID_1 | ID_2 | ID_3 |
+--------+---------+------+------+------+
|    408 | 4/7/17  |  906 |      |  906 |
|    408 | 4/8/17  |  906 |  799 |  906 |
|    408 | 4/9/17  |  906 |  799 |  906 |
|    408 | 4/10/17 |  906 |  600 |  906 |
|    408 | 4/11/17 |  906 |      |  906 |
|    408 | 4/12/17 |  906 |      |  906 |
|    408 | 4/13/17 |  906 |  799 |  906 |
|    408 | 4/14/17 |  906 |  799 |  906 |
+--------+---------+------+------+------+

我需要输出结果为：

+--------+---------+---------+------+------+------+
| EMP_ID | FROM_DT |  TO_DT  | ID_1 | ID_2 | ID_3 |
+--------+---------+---------+------+------+------+
|    408 | 4/8/17  | 4/9/17  |  906 |  799 |  906 |
|    408 | 4/10/17 | 4/10/17 |  906 |  600 |  906 |
|    408 | 4/13/17 | 4/14/17 |  906 |  799 |  906 |
+--------+---------+---------+------+------+------+

不匹配（ID1＆lt;＆gt; ID2或ID1＆lt;＆gt; ID3）应该具有确切的from和to日期，留下没有不匹配的间隙。

Answer 1

这似乎是间隙和岛屿的变种：

select emp_id, id_1, id_2, id_3, min(dt_1), max(dt_1)
from (select t.*,
             row_number() over (partition by emp_id, id_1, id_2, id_3 order by dt_1) as seqnum
      from t
      where id_2 is not null
     ) t
group by emp_id, id_1, id_2, id_3, (dt_1 - seqnum);