Question

参考以下正确回答的问题：
sql (oracle) counting number of overlapping intervals

在Oracle sql数据库中提供下表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   |
+----+------+-------+------+

我现在想找到重叠区间的数量（包括端点）[开始，停止] n_overlap，以及所有stop的{{1}}值的总和相同的id，即：

name

我尝试了此解决方案，该方法有效：

+----+------+-------+------+-----------+------------+
| id | name | start | stop | n_overlap | sum_stops  |
+----+------+-------+------+-----------+------------+
| 1  |   A  |   1   |  5   |     3     |    19      |
+----+------+-------+------+-----------+------------+
| 2  |   A  |   2   |  6   |     3     |    19      |
+----+------+-------+------+-----------+------------+
| 3  |   A  |   4   |  8   |     3     |    19      |
+----+------+-------+------+-----------+------------+
| 4  |   A  |   9   |  10  |     1     |    10      |
+----+------+-------+------+-----------+------------+
| 5  |   B  |   3   |  6   |     2     |    14      |
+----+------+-------+------+-----------+------------+
| 6  |   B  |   4   |  8   |     2     |    14      |
+----+------+-------+------+-----------+------------+
| 7  |   B  |   1   |  2   |     1     |     2      |
+----+------+-------+------+-----------+------------+

但是，有没有一种方法可以压缩两个select / where查询，例如：

select t.*,
   (select count(*)
    from test t2
    where t2.name = t.name and
          t2.start <= t.stop and
          t2.stop >= t.start
   ) as n_overlap,
   (select sum(stop)
    from test t2
    where t2.name = t.name and
          t2.start <= t.stop and
          t2.stop >= t.start
   ) as sum_stops 
from test t;

这会引发select t.*, (select count(*) as n_overlap, sum(stop) as sum_stops from test t2 where t2.name = t.name and t2.start <= t.stop and t2.stop >= t.start ) from test t;错误？

Answer 1

您应该可以使用{this.props.categoryName}和JOIN做自己想做的事情：

GROUP BY

输出：

SELECT t.id, t.name, t.start, t.stop, COUNT(t2.name) AS n_overlap, SUM(t2.stop) AS sum_stops
FROM test t
LEFT JOIN test t2 ON t2.name = t.name AND t2.start <= t.stop AND t2.stop >= t.start
GROUP BY t.id, t.name, t.start, t.stop

Answer 2

这应该做到：

SELECT test.*, n_overlap, sum_stops
FROM test
LEFT JOIN (
    SELECT m.id, COUNT(o.id) AS n_overlap, SUM(o.stop) AS sum_stops
    FROM test AS m
    INNER JOIN test AS o ON
        /* m.id <> o.id AND */
        m.name = o.name AND
        m.stop >= o.start AND
        o.stop >= m.start
    GROUP BY m.id
) AS sq ON test.id = sq.id

我可能会补充说，您的预期输出将考虑任何给定的行与自身重叠（请参阅第4行）。您可能要排除匹配的行。

相同的select / where查询中的sql（oracle）计数和总和

2 个答案: