sql (oracle) 在同一个 select/where 查询中计数和求和答案

【问题标题】：sql (oracle) count and sum within the same select/where querysql (oracle) 在同一个 select/where 查询中计数和求和
【发布时间】：2019-03-04 12:22:15
【问题描述】：

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

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   |
+----+------+-------+------+

我现在想找出重叠间隔的数量（包括端点）[开始，停止]n_overlap 以及所有id 具有相同name 的stop 值的总和，即：

+----+------+-------+------+-----------+------------+
| 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      |
+----+------+-------+------+-----------+------------+

我尝试了这个解决方案，它有效：

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;

但是，有没有办法压缩两个 select/where 查询，使用例如：

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;

会引发too many values 错误？

【问题讨论】：

标签： sql oracle select count where

【解决方案1】：

你应该可以用JOIN 和GROUP BY 做你想做的事：

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

输出：

id  name    start   stop    n_overlap   sum_stops
1   A       1       5       3           19
2   A       2       6       3           19
3   A       5       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

【讨论】：

谢谢@Nick。它给出了一个错误not a GROUP BY expression，但如果我从SELECT 中删除t.* 就可以了
对不起，习惯了 MySQL 松懈的GROUP BY 规则。如果您想要所有 t.* 值（不仅仅是 id），您应该能够将它们全部放在 GROUP BY 例如GROUP BY t.id, t.name, t.start, t.stop
谢谢@Nick。如果我明确列出SELECT 和GROUP BY 中的所有列/值，它会起作用，但如果我只是使用SELECT t.*, COUNT(t2.name) AS... 和GROUP BY t.*，它不会，当有很多列/值时，这可能是一个问题.
@MarcoC 我已经编辑了我的答案以反映这一点。正如您所说，当有很多列和值时，这很痛苦。解决这个问题的唯一方法是使用 Salman 在他的回答中提出的子查询。您可能想接受这一点。

【解决方案2】：

应该这样做：

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

我可能会补充一点，您的预期输出会考虑任何给定的行与其自身重叠（参见第 4 行）。您可能希望排除匹配自身的行。

【讨论】：