SQL not a single-group group function

问题:

When I run the following SQL statement:

SELECT MAX(SUM(TIME))
FROM downloads
GROUP BY SSN

It returns the maximum sum value of downloads by a customer, however if I try to find the social security number that that max value belongs to by adding it to the select statement:

SELECT SSN, MAX(SUM(TIME))
FROM downloads
GROUP BY SSN

I get the following error:

not a single-group group function

I do not understand why it is throwing this error. A google search came up with the following action:

Drop either the group function or the individual column expression from the SELECT list or add a GROUP BY clause that includes all individual column expressions listed

From what I think this is saying - dropping the group function makes the sum value invalid - droping the individual column expression (SSN) will just give me the max sum - not sure about that third part.

Could anyone guide in the right direction?

-Tomek

EDIT: TIME in this database refers to the number of times downloaded

回答1:

Well the problem simply-put is that the SUM(TIME) for a specific SSN on your query is a single value, so it's objecting to MAX as it makes no sense (The maximum of a single value is meaningless).

Not sure what SQL database server you're using but I suspect you want a query more like this (Written with a MSSQL background - may need some translating to the sql server you're using):

SELECT TOP 1 SSN, SUM(TIME)
FROM downloads
GROUP BY SSN
ORDER BY 2 DESC

This will give you the SSN with the highest total time and the total time for it.

Edit - If you have multiple with an equal time and want them all you would use:

SELECT
SSN, SUM(TIME)
FROM downloads
GROUP BY SSN
HAVING SUM(TIME)=(SELECT MAX(SUM(TIME)) FROM downloads GROUP BY SSN))

回答2:

If you want downloads number for each customer, use:

select ssn
     , sum(time)
  from downloads
 group by ssn

If you want just one record -- for a customer with highest number of downloads -- use:

select *
  from (
        select ssn
             , sum(time)
          from downloads
         group by ssn
         order by sum(time) desc
       )
 where rownum = 1

However if you want to see all customers with the same number of downloads, which share the highest position, use:

select *
  from (
        select ssn
             , sum(time)
             , dense_rank() over (order by sum(time) desc) r
          from downloads
         group by ssn
       )
 where r = 1

回答3:

Maybe you find this simpler

select * from (
    select ssn, sum(time) from downloads
    group by ssn
    order by sum(time) desc
) where rownum <= 10 --top 10 downloaders

Regards
K