CareerCup

select customer_id, sum(expense) as se
from table
group by customer_id
order by se desc
limit 5

select customer_id
from (select customer_id,sum(expenses) as exp
from table_name
group by customer_id,product_id)
orderby exp
where rownum<=5

select customer_id, sum(expenses) as sum_exp where rownum <= 5 group by customer_id order by sum_exp desc;

rownum is checked after rows have been fetched.

select top 5
customerid
, sum(expenses) as ep
from testexpense
group by customerid
order by ep desc

select customer_id,sum(expenses) from Customer
group by customer_id
order by sum(expenses) desc

select top 5 CustomerId,Sum(Expenses) from expenses group by CustomerId order by SUM(Expenses) desc

select top 5 CustomerId,Sum(Expenses) from expenses group by CustomerId order by SUM(Expenses) desc

In MySQL:

set @currRank := 0;
select * from (
select customerid, total_exp, @currRank := @currRank + 1 as exp_rank from
(select customerid, sum(totalexpense) as total_exp from expenses
group by customerid) X order by total_exp desc) Y where exp_rank <= 5

SELECT customer_id,SUM(expenses) AS s_exp FROM Customer GROUP BY customer_id ORDER BY s_exp DESC LIMIT 5;

select * from (select custid,sum(exp) from table group by custid) where rownum<=5;

select distinct B.customer_id from (
select A.customer_id,rank(A.total_expenses) over(order by A.total_expenses desc) as Ranks from (
select customer_id,sum(expenses) as total_expenses from customer
group by customer_id ) A ) B
where B.Ranks <=5