More concise SQL query involving MAX() - mysql

Question

inventory
+------------------+-------------------+------------+
| DVD | replacement_price | stock |
+------------------+-------------------+------------+
| Pi | 9.99 | 500 |
| Dune | 29.99 | 100 |
| Heathers | 4.99 | 20 |
| Jaws | 19.99 | 500 |
| Mulholland_Drive | 39.99 | 50 |
| Waking_Life | 29.99 | 200 |
+------------------+-------------------+------------+
rented
+-----------------+-----------+------------------+
| subscriber | queue_nbr | DVD |
+-----------------+-----------+------------------+
| Bob | 1 | Mulholland_Drive |
| Bob | 2 | Jaws |
| Chey | 1 | Pi |
| Chey | 2 | Heathers |
| Jamie | 2 | Mulholland_Drive |
| Jamie | 4 | Dune |
| Jamie | 1 | Jaws |
| Jamie | 3 | Waking_Life |
| Nora | 4 | Jaws |
| Nora | 2 | Mulholland_Drive |
| Nora | 3 | Dune |
| Nora | 1 | Waking_Life |
+-----------------+-----------+------------------+
I want to return ONLY the subscriber(s) with the priciest movie queue (think Netflix DVD replacement costs if you lost all the movies you had out at a given time). I've used MAX() rather than TOP, LIMIT or ROWNUM because the query needs to be as db-independent as possible and must return multiple subscribers in the event of a tie. Using the tables above, the result should be
+---------+
| highest |
+---------+
| Jamie |
| Nora |
+---------+
After much searching and experimentation, I've come up with code that works, but it seems to my novice eyes bloated and inefficient, both in quantity of code and execution.
Would anyone mind refactoring and explaining your code?
My code:
SELECT z.subscriber highest
FROM
(SELECT MAX(price) max_price
FROM (
SELECT subscriber_name subscriber, SUM(replacement_price) price
FROM inventory i
INNER JOIN rented r
ON i.DVD = r.DVD
GROUP BY subscriber
) x
) y
INNER JOIN
(
SELECT subscriber_name subscriber, SUM(replacement_price) price
FROM inventory i
INNER JOIN rented r
ON i.DVD = r.DVD
GROUP BY subscriber
) z
ON z.price = y.max_price

Answer 1

If you want to return only those with the max total, then you could use the following which works in both MySQL and SQL Server. It is not any more concise than your current query though:
select subscriber
from inventory i
inner join rented r
on i.dvd = r.dvd
group by subscriber
having sum(replacement_price) = (select max(TotalCost)
from
(
select sum(replacement_price) TotalCost
from inventory i
inner join rented r
on i.dvd = r.dvd
group by subscriber
) p);
If you are using SQL Server, then I would suggest implementing windowing functions, similar to this:
select subscriber
from
(
select subscriber,
rank() over(order by sum(replacement_price) desc) rnk
from inventory i
inner join rented r
on i.dvd = r.dvd
group by subscriber
) src
where rnk = 1
See SQL Fiddle with Demo

Answer 2

SELECT z.subscriber
FROM(
SELECT RANK() OVER(ORDER BY SUM(replacement_price)) subscriber_rank,
r.subscriber subscriber,
SUM(replacement_price) totalReplacementPrice
FROM inventory i
INNER JOIN rented r ON i.dvd = r.DVD
GROUP BY subscriber
) z
WHERE z.subscriber_rank = 1
Some of your column names are different in you query from you sql sample, so I've used the column names given in the demo tables. I use the rank function in the inner query to find the order of all of the people ordering by the sum of the replacement_price. Then select the row(s) where the rank is 1.
Rank is available in both MS Sql Server and Oracle. To go much further than that as #bluefeet says you will need to give more detail as to which database you are targetting.