Question

我正在尝试从我们的数据库中查找SIP线路上的最大呼叫数。这是数据示例：

ID  StartTime           EndTime
66  01/03/2018 5:31:18  01/03/2018 5:31:51
66  01/03/2018 5:31:51  01/03/2018 5:32:19
66  01/03/2018 5:32:21  01/03/2018 5:34:10
66  01/03/2018 5:32:31  01/03/2018 5:33:42
66  01/03/2018 5:33:02  01/03/2018 5:33:42
66  01/03/2018 5:33:23  01/03/2018 5:34:10
66  01/03/2018 5:33:45  01/03/2018 5:34:38
66  01/03/2018 5:34:08  01/03/2018 5:34:38

从视觉上看，我的计算方法应该是5。

到目前为止，我所得出的最接近值在该示例中返回了4，并且在其他测试中也处于关闭状态。我测试过的其他代码异常繁琐（100秒）。

DECLARE @ReportStartDate datetime, @ReportEndDate datetime,        
@ReportStartTime time, @ReportEndTime time
SET @ReportStartDate='1/3/2018'
SET @ReportEndDate='1/3/2018'
SET @ReportStartTime='05:00:00'
SET @ReportEndTime='06:00:00'

SET @ReportStartDate = (SELECT @ReportStartDate + @ReportStartTime)
SET @ReportEndDate = (SELECT @ReportEndDate + @ReportEndTime);

WITH C1 AS
(
  SELECT startdate AS ts, +1 AS TYPE,
    ROW_NUMBER() OVER(ORDER BY startdate) AS start_ordinal
  FROM table
  WHERE startdate >= @ReportStartDate and startdate <= @ReportEndDate
  and Service_id=66 --hardcoded for testing
  UNION ALL
  SELECT enddate, -1, NULL
  FROM table
  WHERE enddate >= @ReportStartDate and enddate <=    
 @ReportEndDate
  and Service_id=66 -- hardcoded for testing
),
C2 AS
(
  SELECT *,
    ROW_NUMBER() OVER(  ORDER BY ts, TYPE) AS start_or_end_ordinal
  FROM C1
)
SELECT MAX(2 * start_ordinal - start_or_end_ordinal) AS mx
FROM C2
WHERE TYPE = 1

Answer 1

我将此逻辑写为：

with tt as (
      select id, stime as thetime, 1 as inc from t union all
      select id, etime, -1 as inc from t
     )
select top (1) tt.*
from (select tt.*, sum(inc) over (partition by id order by thetime) as overlaps
      from tt
     ) tt
order by overlaps desc;

对于您的示例，这也会返回“ 4”，但这对我来说很有意义。

您可以在this db<>fiddle上进行测试。

从数据库获取最大同时呼叫次数

1 个答案: