SQL - Finding Customer's largest Location by Order $ - sql

Question

I have a table with customer IDs, location IDs, and their order values. I need to select the location ID for each customer with the largest spend
Customer | Location | Order $
1 | 1A | 100
1 | 1A | 20
1 | 1B | 100
2 | 2A | 50
2 | 2B | 20
2 | 2B | 50
So I would get
Customer | Location | Order $
1 | 1A | 120
2 | 2B | 70
I tried something like this:
SELECT
a.CUST
,a.LOC
,c.BOOKINGS
FROM (SELECT DISTINCT TOP 1 b.CUST, b.LOC, sum(b.ORDER_VAL) as BOOKINGS
FROM ORDER_TABLE b
GROUP BY b.CUST, b.LOC
ORDER BY BOOKINGS DESC) as c
INNER JOIN ORDER_TABLE a
ON a.CUST = c.CUST
But that just returns the top order.

Answer 1

Just use variables to emulate ROW_NUM()
DEMO
SELECT *
FROM ( SELECT `Customer`, `Location`, SUM(`Order`) as `Order`,
#rn := IF(#customer = `Customer`,
#rn + 1,
IF(#customer := `Customer`, 1, 1)
) as rn
FROM Table1
CROSS JOIN (SELECT #rn := 0, #customer := '') as par
GROUP BY `Customer`, `Location`
ORDER BY `Customer`, SUM(`Order`) DESC
) t
WHERE t.rn = 1

Answer 2

Firs you have to sum the values for each location:
select Customer, Location, Sum(Order) as tot_order
from order_table
group by Customer, Location
then you can get the maximum order with MAX, and the top location with a combination of group_concat that will return all locations, ordered by total desc, and substring_index in order to get only the top one:
select
Customer,
substring_index(
group_concat(Location order by tot_order desc),
',', 1
) as location,
Max(tot_order) as max_order
from (
select Customer, Location, Sum(Order) as tot_order
from order_table
group by Customer, Location
) s
group by Customer
(if there's a tie, two locations with the same top order, this query will return just one)

Answer 3

This seems like an order by using aggregate function problem. Here is my stab at it;
SELECT
c.customer,
c.location,
SUM(`order`) as `order_total`,
(
SELECT
SUM(`order`) as `order_total`
FROM customer cm
WHERE cm.customer = c.customer
GROUP BY location
ORDER BY `order_total` DESC LIMIT 1
) as max_order_amount
FROM customer c
GROUP BY location
HAVING max_order_amount = order_total
Here is the SQL fiddle. http://sqlfiddle.com/#!9/2ac0d1/1

Answer 4

This is how I'd handle it (maybe not the best method?) - I wrote it using a CTE first, only to see that MySQL doesn't support CTEs, then switched to writing the same subquery twice:
SELECT B.Customer, C.Location, B.MaxOrderTotal
FROM
(
SELECT A.Customer, MAX(A.OrderTotal) AS MaxOrderTotal
FROM
(
SELECT Customer, Location, SUM(`Order`) AS OrderTotal
FROM Table1
GROUP BY Customer, Location
) AS A
GROUP BY A.Customer
) AS B INNER JOIN
(
SELECT Customer, Location, SUM(`Order`) AS OrderTotal
FROM Table1
GROUP BY Customer, Location
) AS C ON B.Customer = C.Customer AND B.MaxOrderTotal = C.OrderTotal;
Edit: used the table structure provided
This solution will provide multiple rows in the event of a tie.
SQL fiddle for this solution

Answer 5

How about:
select a.*
from (
select customer, location, SUM(val) as s
from orders
group by customer, location
) as a
left join
(
select customer, MAX(b.tot) as t
from (
select customer, location, SUM(val) as tot
from orders
group by customer, location
) as b
group by customer
) as c
on a.customer = c.customer where a.s = c.t;

Answer 6

with
Q_1 as
(
select customer,location, sum(order_$) as order_sum
from cust_order
group by customer,location
order by customer, order_sum desc
),
Q_2 as
(
select customer,max(order_sum) as order_max
from Q_1
group by customer
),
Q_3 as
(
select Q_1.customer,Q_1.location,Q_1.order_sum
from Q_1 inner join Q_2 on Q_1.customer = Q_2.customer and Q_1.order_sum = Q_2.order_max
)
select * from Q_3
Q_1 - selects normal aggregate, Q_2 - selects max(aggregate) out of Q_1 and Q_3 selects customer,location, sum(order) from Q_1 which matches with Q_2