mysql查询 - 峰值并发调用CDR数据

Question

嗨，我需要帮助解决如何计算我存储在MySQL中的CDR日期中一天中有多少峰值并发呼叫。

数据集如下所示：

INSERT INTO `cdr` (`calldate`, `clid`, `src`, `dst`, `dcontext`, `channel`, `dstchannel`, 
                   `lastapp`, `lastdata`, `duration`, `billsec`, `disposition`, `amaflags`, 
                   `accountcode`, `uniqueid`, `userfield`) VALUES

我可以使用以下查询来计算每个唯一日期的条目数。

SELECT COUNT(1) AS entries, date(calldate) AS DATE
FROM  `cdr` 
GROUP BY DATE (calldate)
LIMIT 0 , 1000

然而，这只告诉我并发呼叫的理论峰值而不是实际峰值。

要获得实际的峰值，我们需要先了解每次通话的开始和结束日期和时间。 当前开始日期和时间记录在（校准）字段中，并且呼叫的持续时间以秒为单位记录在（持续时间）字段中。 通过将（持续时间）字段中存储的秒添加到（calldate）字段，我们可以计算完成时间。

现在我们知道了开始和结束时间，我们需要计算这些时间是否重叠以及重复多少次。 这种级别的SQL查询远远超出我的知识。

回顾一下，我试图用MySQL查询计算同时调用的峰值数来自MySQL中存储的CDR数据。 感谢任何帮助

样本数据：

    calldate,clid,src,dst,dcontext,channel,dstchannel,lastapp,lastdata,duration,billsec,disposition,amaflags,accountcode,uniqueid,userfield
08/11/2013 17:02,x,x,1000,default,x,x,x,x,26,26,ANSWERED,3,x,1383930162,x
08/11/2013 17:02,x,x,1000,default,x,x,x,x,24,24,ANSWERED,3,x,1383930164,x
08/11/2013 17:02,x,x,1000,default,x,x,x,x,45,45,ANSWERED,3,x,1383930146,x
08/11/2013 17:10,x,x,1000,default,x,x,x,x,2,2,ANSWERED,3,x,1383930649,x
08/11/2013 17:22,x,x,1000,default,x,x,x,x,4,4,ANSWERED,3,x,1383931380,x
08/11/2013 17:23,x,x,1000,default,x,x,x,x,5,5,ANSWERED,3,x,1383931388,x
08/11/2013 17:23,x,x,1000,default,x,x,x,x,9,9,ANSWERED,3,x,1383931395,x
10/11/2013 09:28,x,x,1000,default,x,x,x,x,7,7,ANSWERED,3,x,1384075689,x
10/11/2013 10:09,x,x,1000,default,x,x,x,x,57,57,ANSWERED,3,x,1384078181,x
10/11/2013 10:09,x,x,1000,default,x,x,x,x,81,81,ANSWERED,3,x,1384078164,x
10/11/2013 10:09,x,x,1000,default,x,x,x,x,102,102,ANSWERED,3,x,1384078143,x
11/11/2013 10:23,x,x,1000,default,x,x,x,x,3,3,ANSWERED,3,x,1384165439,x
11/11/2013 17:46,x,x,1000,default,x,x,x,x,30,30,ANSWERED,3,x,1384191975,x
11/11/2013 17:46,x,x,1000,default,x,x,x,x,30,30,ANSWERED,3,x,1384191976,x
11/11/2013 17:45,x,x,1000,default,x,x,x,x,50,50,ANSWERED,3,x,1384191956,x
11/11/2013 17:55,x,x,1000,default,x,x,x,x,9,9,ANSWERED,3,x,1384192544,x
13/11/2013 10:59,x,x,1000,default,x,x,x,x,209,209,ANSWERED,3,x,1384340382,x
13/11/2013 10:59,x,x,1000,default,x,x,x,x,230,230,ANSWERED,3,x,1384340361,x
13/11/2013 11:09,x,x,1000,default,x,x,x,x,1342,1342,ANSWERED,3,x,1384340963,x
13/11/2013 11:10,x,x,1000,default,x,x,x,x,1312,1312,ANSWERED,3,x,1384341009,x
13/11/2013 11:08,x,x,1000,default,x,x,x,x,1441,1441,ANSWERED,3,x,1384340891,x
13/11/2013 11:10,x,x,1000,default,x,x,x,x,1288,1288,ANSWERED,3,x,1384341059,x
13/11/2013 11:10,x,x,1000,default,x,x,x,x,1306,1306,ANSWERED,3,x,1384341050,x
13/11/2013 11:09,x,x,1000,default,x,x,x,x,1378,1378,ANSWERED,3,x,1384340990,x
13/11/2013 11:09,x,x,1000,default,x,x,x,x,1419,1419,ANSWERED,3,x,1384340953,x
13/11/2013 11:06,x,x,1000,default,x,x,x,x,1558,1558,ANSWERED,3,x,1384340815,x
13/11/2013 11:12,x,x,1000,default,x,x,x,x,1254,1254,ANSWERED,3,x,1384341121,x
13/11/2013 11:10,x,x,1000,default,x,x,x,x,1330,1330,ANSWERED,3,x,1384341045,x
13/11/2013 11:09,x,x,1000,default,x,x,x,x,1431,1431,ANSWERED,3,x,1384340947,x
13/11/2013 11:11,x,x,1000,default,x,x,x,x,1302,1302,ANSWERED,3,x,1384341076,x
13/11/2013 11:09,x,x,1000,default,x,x,x,x,1383,1383,ANSWERED,3,x,1384340995,x
13/11/2013 11:08,x,x,1000,default,x,x,x,x,1444,1444,ANSWERED,3,x,1384340937,x
13/11/2013 11:07,x,x,1000,default,x,x,x,x,1531,1531,ANSWERED,3,x,1384340850,x
13/11/2013 11:09,x,x,1000,default,x,x,x,x,1418,1418,ANSWERED,3,x,1384340963,x
13/11/2013 12:02,x,x,1000,default,x,x,x,x,10,10,ANSWERED,3,x,1384344169,x
13/11/2013 12:01,x,x,1000,default,x,x,x,x,807,807,ANSWERED,3,x,1384344072,x
13/11/2013 12:03,x,x,1000,default,x,x,x,x,680,680,ANSWERED,3,x,1384344200,x
13/11/2013 12:01,x,x,1000,default,x,x,x,x,793,793,ANSWERED,3,x,1384344090,x
13/11/2013 12:01,x,x,1000,default,x,x,x,x,772,772,ANSWERED,3,x,1384344111,x

Answer 1

这应该工作，但是真正的性能杀手！

SELECT
  calldate,
  MAX(concurrent)+1 AS peakcount
FROM (
    SELECT
      DATE(a.calldate) as calldate,
      COUNT(b.uniqueid) AS concurrent
    FROM cdr AS a, cdr AS b
    WHERE  
      a.calldate BETWEEN '2013-11-08 00:00:00' AND '2013-11-13 23:59:59'
      AND (
        (a.calldate<=b.calldate AND (UNIX_TIMESTAMP(a.calldate)+a.duration)>=UNIX_TIMESTAMP(b.calldate))
        OR (b.calldate<=a.calldate AND (UNIX_TIMESTAMP(b.calldate)+b.duration)>=UNIX_TIMESTAMP(a.calldate))
      )
      AND a.uniqueid>b.uniqueid
    GROUP BY a.uniqueid
  ) AS baseview
GROUP BY calldate

为您的示例数据提供正确的答案。 下面是它的工作原理：

• 最里面的部分（ a.calldate<=b.calldate AND (UNIX_TIMESTAMP(a.calldate)+a.duration)>=UNIX_TIMESTAMP(b.calldate) ...）计算交集：两个调用重叠，如果起点为一个呼叫在另一个呼叫的起始点或之后以及该呼叫的结束点或之前
• 自动加入调用表可查找所有重叠，
• 但有一个问题：自联接在第1行和第2行之间找到了一圈，但另一行在第2行和第1行之间找到了一圈。如果两个以上的调用重叠，那么将它排除在外是很繁琐的。
• 现在，由于您的数据包含数字唯一ID，我们可以使用它来过滤那些重复，重复等等。这是由AND a.uniqueid>b.uniqueid选择器和GROUP BY a.uniqueid ，它只使用最小的uniqueid看到所有并发呼叫，其他人看得少
• 在外部查询中使用MAX()过滤掉此记录
• 我们需要+1来获得峰值呼叫计数：具有2个并发呼叫的呼叫意味着峰值计数为3

SQLfiddle

Answer 2

尝试为每个调用开始添加+1，为每个结尾添加-1，然后只获得该+ 1 / -1列的累积和
***如果需要，转换calldate或使用您的格式：

set @from:='2015-02-01';
set @to:='2015-03-01';
set @csum:=0;
SELECT DT,CallCount, (@csum := @csum + CallCount) as cumulative_sum
FROM 
(select calldate AS DT, 1 AS CallCount
 from cdr
 where calldate between @from and @to

 union all

 select ADDDATE(calldate,INTERVAL duration SECOND) AS DT, -1 AS CallCount
 from cdr 
 where calldate between @from and @to
  ) Calls
ORDER BY 1 asc;

Answer 3

使用组，让交换机将总呼叫数+1写入新的CDR值以实现此目的。 所以你知道最后一次呼叫进入系统时有多少通话。