SQL对具有重叠时间跨度的数据进行分组

我需要根据记录开始和结束时间通过重叠时间跨度将彼此相关的数据分组在一起。 SQL-fiddle在这里：http://sqlfiddle.com/#!18/87e4b/1/0 http://sqlfiddle.com/#!18/87e4b/1/0

我构建的当前查询给出了不正确的结果。 Callid 3 的 callCount 应该为 4。这不是因为记录 6 不包含在内，因为它与 3 不重叠，但应该包含在内，因为它确实与其他相关记录之一重叠。所以我相信可能需要递归 CTE，但我不确定如何编写它。

Schema:

CREATE TABLE Calls
    ([callid] int, [src] varchar(10), [start] datetime, [end] datetime, [conf] varchar(5));

INSERT INTO Calls
    ([callid],[src],[start],[end],[conf])
VALUES
    ('1','5555550001','2019-07-09 10:00:00', '2019-07-09 10:10:00', '111'),
    ('2','5555550002','2019-07-09 10:00:01', '2019-07-09 10:11:00', '111'),
    ('3','5555550011','2019-07-09 11:00:00', '2019-07-09 11:10:00', '111'),
    ('4','5555550012','2019-07-09 11:00:01', '2019-07-09 11:11:00', '111'),
    ('5','5555550013','2019-07-09 11:01:00', '2019-07-09 11:15:00', '111'),
    ('6','5555550014','2019-07-09 11:12:00', '2019-07-09 11:16:00', '111'),
    ('7','5555550014','2019-07-09 15:00:00', '2019-07-09 15:01:00', '111');

当前查询：

SELECT 
    detail_record.callid,
    detail_record.conf,
    MIN(related_record.start) AS sessionStart,
    MAX(related_record.[end]) As sessionEnd,
    COUNT(related_record.callid) AS callCount
FROM    
    Calls AS detail_record
    INNER JOIN
    Calls AS related_record     
        ON related_record.conf = detail_record.conf
        AND ((related_record.start >= detail_record.start
                AND related_record.start < detail_record.[end])
            OR (related_record.[end] > detail_record.start
                AND related_record.[end] <= detail_record.[end])
            OR (related_record.start <= detail_record.start
                AND related_record.[end] >= detail_record.[end])
            )
WHERE
    detail_record.start > '1/1/2019'
    AND detail_record.conf = '111'
GROUP BY
    detail_record.callid,
    detail_record.start,
    detail_record.conf
HAVING 
    MIN(related_record.start) >= detail_record.start
ORDER BY sessionStart DESC

预期成绩：

callid  conf  sessionStart          sessionEnd              callCount
   7    111   2019-07-09T15:00:00Z  2019-07-09T15:01:00Z    1
   3    111   2019-07-09T11:00:00Z  2019-07-09T11:15:00Z    4
   1    111   2019-07-09T10:00:00Z  2019-07-09T10:11:00Z    2

这是一个缺口和岛屿问题。它不需要递归 CTE。您可以使用窗口函数：

select min(callid), conf, grouping, min([start]), max([end]), count(*)
from (select c.*,
             sum(case when prev_end < [start] then 1 else 0 end) over (order by start) as grouping
      from (select c.*,
                   max([end]) over (partition by conf order by [start] rows between unbounded preceding and 1 preceding) as prev_end
            from calls c
           ) c
     ) c
group by conf, grouping;

最里面的子查询计算前一个结尾。中间子查询将其与当前开始进行比较，以确定相邻行组何时成为新组的开始。然后通过累积和确定分组。

并且，外部查询聚合以汇总有关每个组的信息。

是一个数据库小提琴。