SQL question for to get a top customer

Question

Hi All,
I am prasanna. I have a question on SQL,to get a top customer based on year and customer name.I written query like this. But i got an error.
select calendar_year,cust_first_name,max(sum(amount_sold)) from sales,times,customers where sales.cust_id=customers.cust_id and times.time_id=sales.time_id group  by calendar_year,cust_first_name
The error is like this:
*Error starting at line 1 in command:
select calendar_year,cust_first_name,max(sum(amount_sold)) from sales,times,customers where sales.cust_id=customers.cust_id and times.time_id=sales.time_id group by calendar_year,cust_first_name
Error at Command Line:1 Column:7
Error report:
SQL Error: ORA-00937: not a single-group group function
00937. 00000 - "not a single-group group function"*
*Cause:   
Action:
Thanks inadvace
Regards,
prasanna

Answer 1

It is not clear what you want. Code below will return a customer with highest total amout sold regardless of year:
select  calendar_year,
        cust_first_name,
        total_amount_sold max_amount_sold
  from  (
         select  calendar_year,
                 cust_first_name,
                 sum(amount_sold) total_amount_sold,
                 row_number() over(order by sum(amount_sold) desc) rn
           from  sales,
                 times,
                 customers
           where sales.cust_id=customers.cust_id
             and times.time_id=sales.time_id
           group by calendar_year,
                 cust_first_name
  where rn = 1
/Keep in mind, if more than one customer has that highest total amount sold and you want all such customers:
select  calendar_year,
        cust_first_name,
        total_amount_sold max_amount_sold
  from  (
         select  calendar_year,
                 cust_first_name,
                 sum(amount_sold) total_amount_sold,
                 rank() over(order by sum(amount_sold) desc) rn
           from  sales,
                 times,
                 customers
           where sales.cust_id=customers.cust_id
             and times.time_id=sales.time_id
           group by calendar_year,
                 cust_first_name
  where rn = 1
/And if you want top customer for each calendar year:
select  calendar_year,
        cust_first_name,
        total_amount_sold max_amount_sold
  from  (
         select  calendar_year,
                 cust_first_name,
                 sum(amount_sold) total_amount_sold,
                 row_number() over(partition by calendar_year order by sum(amount_sold) desc) rn
           from  sales,
                 times,
                 customers
           where sales.cust_id=customers.cust_id
             and times.time_id=sales.time_id
           group by calendar_year,
                 cust_first_name
  where rn = 1
/Keep in mind, if more than one have that highest total amount sold and you want all such customers:
select  calendar_year,
        cust_first_name,
        total_amount_sold max_amount_sold
  from  (
         select  calendar_year,
                 cust_first_name,
                 sum(amount_sold) total_amount_sold,
                 rank() over(partition by calendar_year order by sum(amount_sold) desc) rn
           from  sales,
                 times,
                 customers
           where sales.cust_id=customers.cust_id
             and times.time_id=sales.time_id
           group by calendar_year,
                 cust_first_name
  where rn = 1
/SY.
Edited by: Solomon Yakobson on Dec 22, 2011 9:24 AM