mysql query - peak concurrent calls CDR data

问题

Hi I need help working out how to calculate how many peak concurrent calls I have in a day from my CDR date stored in MySQL.

The data set looks like this:

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

I can user the following query to calculate how many entries there are for each unique date.

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

However this only tells me the theoretical peak of concurrent calls are and not the actual peak.

To get the actual peak we need to first know the start and finish dates and times of each call. Currently the start date and time is recorded in the (calldate) field and the duration of the call is recorded in the (duration) field in seconds. By adding the seconds stored in the (duration) field to the (calldate) field we are able to calculate the finish time.

Now that we know the start and finish times we need to calculate if these times overlap and by how many times. This level of SQL query is well beyond my knowledge.

To recap I am trying to calculate with a MySQL query what the peak number of simultaneous calls are from the CDR data stored in MySQL. Any help is gratefully received

Sample Data:

    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

回答1:

This one should work, but is a real performance killer!

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

gives the correct answers for your example data. Here is, how it works:

The innermost part (a.calldate<=b.calldate AND (UNIX_TIMESTAMP(a.calldate)+a.duration)>=UNIX_TIMESTAMP(b.calldate)...) calculates the intersection: Two calls overlap, if the starting point of one call is at or after the starting point of the other call and at or before the ending point of that call
Self-joining the call tables finds all overlaps,
but with a problem: The self join finds an over lap between lines 1 and 2, but another one with lines 2 and 1. If more than two calls overlap, it is tedious to sort this out
Now since your data contains a numeric unique ID, we can use this to filter those duplicates, triplicates etc. this is done by the AND a.uniqueid>b.uniqueid selector and GROUP BY a.uniqueid, which makes only the call with the smallest uniqueid see all concurrent calls, the others see less
Using MAX() on this in the outer query filters out this record
We need the +1 to get the peak call count: A call with 2 concurrent calls means a peak count of 3

SQLfiddle

回答2:

Try adding +1 for each start of call, and -1 for each end, then just get cumulative sum of that +1/-1 column
***Convert the calldate if needed or use your format:

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;

回答3:

Using Groups, have the switch write the total calls up +1 to a new CDR value for such purpose. so you know how many calls were up when the last call entered the system.

来源：https://stackoverflow.com/questions/21420744/mysql-query-peak-concurrent-calls-cdr-data

标签

mysql

concurrency

cdr