sql（oracle）计算重叠间隔的数量

Question

我有以下问题：

给定oracle sql数据库中的下表test：

+----+------+-------+------+
| id | name | start | stop |
+----+------+-------+------+
| 1  |   A  |   1   |  5   |
+----+------+-------+------+
| 2  |   A  |   2   |  6   |
+----+------+-------+------+
| 3  |   A  |   5   |  8   |
+----+------+-------+------+
| 4  |   A  |   9   |  10  |
+----+------+-------+------+
| 5  |   B  |   3   |  6   |
+----+------+-------+------+
| 6  |   B  |   4   |  8   |
+----+------+-------+------+
| 7  |   B  |   1   |  2   |
+----+------+-------+------+

我想找出所有具有相同name 的id 的重叠间隔（包括端点）[开始，停止] n_overlap 的数量，即：

+----+------+-------+------+-----------+
| id | name | start | stop | n_overlap |
+----+------+-------+------+-----------+
| 1  |   A  |   1   |  5   |     3     |
+----+------+-------+------+-----------+
| 2  |   A  |   2   |  6   |     3     |
+----+------+-------+------+-----------+
| 3  |   A  |   4   |  8   |     3     |
+----+------+-------+------+-----------+
| 4  |   A  |   9   |  10  |     1     |
+----+------+-------+------+-----------+
| 5  |   B  |   3   |  6   |     2     |
+----+------+-------+------+-----------+
| 6  |   B  |   4   |  8   |     2     |
+----+------+-------+------+-----------+
| 7  |   B  |   1   |  2   |     1     |
+----+------+-------+------+-----------+

Answer 1

一种方法使用相关子查询：

select t.*,
       (select count(*)
        from test t2
        where t2.name = t.name and
              t2.start < t.end and
              t2.end > t.start
       ) as num_overlaps
from test t;