Question

I’m using SQL Server 2008 With this table :

and i have the following query :

SELECT caller, called, SUM(duration) as duration, 
      dateadd(DAY,0, datediff(day,0, time))as Batch2 ,
      CONVERT(VARCHAR(8), time, 4) AS Batch  
FROM  Calls oc
WHERE caller='somevalue'
GROUP BY called, dateadd(DAY,0, datediff(day,0, time)),
          CONVERT(VARCHAR(8), time, 4),
          caller
ORDER BY duration DESC,called

Now i want to order the query by the amount of rows in the query that contain the same called value, meaning first results will be the called that appeared in most days.

Now the closest i got was using :

order by (select count(called) from Calls where called=oc.called ),
         duration DESC,
         called

But this returns all the appearnces of called in general and not grouped by as my initial query distincts values.

And i can’t use OVER() as that will just return the number of rows in general.

This is an example of the output i have for now :

callervalue   somecalledvalue   589 2012-11-21 00:00:00.000 21.11.12
callervalue   somecalledvalue   551 2012-11-20 00:00:00.000 20.11.12
callervalue   somecalledvalue   506 2012-11-22 00:00:00.000 22.11.12

How can i achive this?

Answer 1

The answer was posted by Chris Bednarski as a comment :

try COUNT(called) OVER (PARTITION BY called)

Which did the trick.