Question

I have this as a answer since earlier today

SELECT  Users.full_name
        ,sum(total)
        ,sum(case when service < 49 then total else 0 end) AS Productive
FROM    time_report
        , users
WHERE   time_report.User = users.user_id
        AND date BETWEEN '2011-0502' AND '2011-05-11'
GROUP BY 
        User

THis gives me the following data back which is what I wanted

full_name   sum(total)  Productive

Cian Higgins    26  14

Wallace Ward    23  23

jason ward      42  33

Thomas Woods    72  53

Peter Jones     49  41

fintan corrigan 40  32

David Jones     35  27

January Jones   23  23

Joe Johnson     24  24

I am wondering it there a way to add an aditional column called percentage to the output which will give me the output from Round((Total/productive) * 100)

I have been struggling with this for most of the day.
I have created a temp table with the output from the original query, but am struggling with the right syntax to get the percentage entered into it.

Any help appreciated.

Answer 1

SELECT  Users.full_name
        ,sum(total)
        ,sum(case when service < 49 then total else 0 end) AS Productive
        ,(Round((sum(total)/sum(case when service < 49 then total else 0 end)) * 100)) AS Percent
...