SQL Query with combining MAX() and SUM() aggregates - sql

Question

I have tried looking into different topics over here and in other forums but I can't seem to find a solution to my problem.
What I'm trying to achieve is "Display the net sales (in dollars) of the Product Line with the highest revenue for that Customer. Use a heading of: Best Sales. Format as $999,999.99.
Here's what I've tried so far:
SELECT cc.CustID, cc.CompanyName, cc.ContactName, pl.pl_id,to_char((sum(od.unitprice*od.quantity*(1-discount))), '$9,999,999.99') as rev
FROM corp.customers cc JOIN corp.orders co ON (cc.CustID=co.CustID)
LEFT OUTER JOIN corp.order_details od ON (co.orderID=od.orderID)
LEFT OUTER JOIN corp.products cp ON (od.ProductID=cp.ProductID)
LEFT OUTER JOIN corp.product_lines pl ON (cp.pl_id=pl.pl_id)
GROUP BY cc.CustID, cc.CompanyName, cc.ContactName, pl.pl_id
HAVING sum(od.unitprice*od.quantity*(1-discount))=
(
SELECT max(sum(od.unitprice*od.quantity*(1-discount)))
FROM corp.customers cc JOIN corp.orders co ON (cc.CustID=co.CustID)
JOIN corp.order_details od ON (co.orderID=od.orderID)
JOIN corp.products cp ON (od.ProductID=cp.ProductID)
JOIN corp.product_lines pl ON (cp.pl_id=pl.pl_id)
GROUP BY cc.CustID, cc.CompanyName, cc.ContactName, pl.pl_id);
This gives me only one output indicating the highest revenue of all customers, but I would like it to display the highest revenue according to each of the product line for that customer.
The result is shown below.
CustID | Company Name | Contact Name | PL_ID | Revenue
QUICK | QUICK-Stop | Horst Kloss | 1 | $37,161.63
I would like it to show something like.
CustID | Company Name | Contact Name | PL_ID | Revenue
QUICK | QUICK-Stop | Horst Kloss | 1 | $37,161.63
QS | QUICK-Start | Clark Stone | 2 | $50,000.00
QUI | QUICK | Mary Haynes | 1 | $60,000.00
QShelf | QUICK-Shelf | Doreen Lucas | 4 | $35,161.63
Any help is appreciated. Thank you!

Answer 1

This query uses your original query, a rank() function to order by your rev column, and a selection to only get the highest rev. This will give multiple rows if you have multiple rows with the same rev value. Change rank() to row_number() if you only want one.
You could also use CTE instead of the nested queries, wont make any difference.
select CustID, CompanyName, ContactName, pl_id, rev from (
select CustID, CompanyName, ContactName, pl_id, to_char(rev, '$9,999,999.99') as rev,
rank() over(order by rev desc) r
from (
SELECT cc.CustID, cc.CompanyName, cc.ContactName, pl.pl_id,
sum(od.unitprice*od.quantity*(1-discount)) as rev
FROM corp.customers cc JOIN corp.orders co ON (cc.CustID=co.CustID)
LEFT OUTER JOIN corp.order_details od ON (co.orderID=od.orderID)
LEFT OUTER JOIN corp.products cp ON (od.ProductID=cp.ProductID)
LEFT OUTER JOIN corp.product_lines pl ON (cp.pl_id=pl.pl_id)
GROUP BY cc.CustID, cc.CompanyName, cc.ContactName, pl.pl_id
) q
) q2 where r=1

Answer 2

Since you didn't provide us with sample input data for your tables, I've knocked up a simple example that you can hopefully use to amend your query:
WITH sample_data AS (SELECT 1 ID, 1 id2, 10 val FROM dual UNION ALL
SELECT 1 ID, 1 id2, 20 val FROM dual UNION ALL
SELECT 1 ID, 2 id2, 30 val FROM dual UNION ALL
SELECT 1 ID, 2 id2, 40 val FROM dual UNION ALL
SELECT 2 ID, 1 id2, 50 val FROM dual UNION ALL
SELECT 2 ID, 2 id2, 60 val FROM dual UNION ALL
SELECT 2 ID, 3 id2, 60 val FROM dual)
SELECT ID,
id2,
max_sum_val
FROM (SELECT ID,
id2,
SUM(val) sum_val,
MAX(SUM(val)) OVER (PARTITION BY ID) max_sum_val
FROM sample_data
GROUP BY ID, id2)
WHERE sum_val = max_sum_val;
ID ID2 MAX_SUM_VAL
---------- ---------- -----------
1 2 70
2 2 60
2 3 60
This will display all id2 values that have the same sum(val) that's the highest. If you don't want to display all tied rows, you can used the row_number() analytic function instead:
WITH sample_data AS (SELECT 1 ID, 1 id2, 10 val FROM dual UNION ALL
SELECT 1 ID, 1 id2, 20 val FROM dual UNION ALL
SELECT 1 ID, 2 id2, 30 val FROM dual UNION ALL
SELECT 1 ID, 2 id2, 40 val FROM dual UNION ALL
SELECT 2 ID, 1 id2, 50 val FROM dual UNION ALL
SELECT 2 ID, 2 id2, 60 val FROM dual UNION ALL
SELECT 2 ID, 3 id2, 60 val FROM dual)
SELECT ID,
id2,
max_sum_val
FROM (SELECT ID,
id2,
SUM(val) sum_val,
row_number() OVER (PARTITION BY ID ORDER BY SUM(val) DESC, id2) rn
FROM sample_data
GROUP BY ID, id2)
WHERE rn = 1;
ID ID2 MAX_SUM_VAL
---------- ---------- -----------
1 2 70
2 2 60
ETA:
That means your query would end up something like:
SELECT custid,
companyname,
contactname,
pl_id,
to_char(rev, '$9,999,999.99') rev
FROM (SELECT cc.custid,
cc.companyname,
cc.contactname,
pl.pl_id,
SUM(od.unitprice * od.quantity * (1 - discount)) AS rev,
MAX(SUM(od.unitprice * od.quantity * (1 - discount))) OVER (PARTITION BY cc.custid) max_rev
FROM corp.customers cc
INNER JOIN corp.orders co ON (cc.custid = co.custid)
LEFT OUTER JOIN corp.order_details od ON (co.orderid = od.orderid)
LEFT OUTER JOIN corp.products cp ON (od.productid = cp.productid)
LEFT OUTER JOIN corp.product_lines PL ON (cp.pl_id = pl.pl_id)
GROUP BY cc.custid,
cc.companyname,
cc.contactname,
pl.pl_id)
WHERE rev = max_rev;