mysql - 最小化 MySQL 查询的冗余

Question

我在尝试减少 MySQL 中的查询冗余时遇到了一些麻烦。我目前正在使用它，但感觉我的开销太多了，因为它使用了冗余子查询。我想做的是使用 dvd 租赁数据库来查找哪个商店位置在 2005 年每个月出租了更多的 dvd。

这是工作查询

SELECT b.month, c.store_id, b.maxRentals
FROM
    (SELECT a.month, MAX(a.rentalCount) as maxRentals
    FROM
        (SELECT MONTH(rental.rental_date) as month, inventory.store_id, count(1) as rentalCount
        FROM rental

        INNER JOIN inventory
        ON rental.inventory_id = inventory.inventory_id

        WHERE YEAR(rental.rental_date) = 2005

        GROUP BY MONTH(rental.rental_date), inventory.store_id
        ) a
    GROUP BY a.month
    ) b

    INNER JOIN

    (SELECT MONTH(rental.rental_date) as month, inventory.store_id, count(1) as rentalCount
    FROM rental

    INNER JOIN inventory
    ON rental.inventory_id = inventory.inventory_id

    WHERE YEAR(rental.rental_date) = 2005

    GROUP BY MONTH(rental.rental_date), inventory.store_id
    ) c
ON b.maxRentals = c.rentalCount

GROUP BY b.month;

注意别名“c”的子查询与别名“a”的子查询完全相同。我不确定是否有办法摆脱这种情况，因为我无法在内部加入别名。我只是被一个巨大的查询困住了，还是我还能做些什么？

Answer 1

我 90% 确定此查询将实现您的意图:

SELECT MONTH(r.rental_date), i.store_id, COUNT(*)
FROM rental r
LEFT JOIN inventory i ON r.inventory_id = i.inventory_id
WHERE YEAR(r.rental_date) = 2005
GROUP BY MONTH(r.rental_date), i.store_id

让我知道进展如何!

编辑:回答2005 年哪个商店位置每个月出租的 dvd 更多的问题:

SELECT x.rental_month, x.store_id, MAX(x.rental_count) FROM (
SELECT MONTH(r.rental_date) AS rental_month, i.store_id AS store_id, COUNT(*) AS rental_count
FROM rental r LEFT JOIN inventory i ON r.inventory_id = i.inventory_id
WHERE YEAR(r.rental_date) = 2005
GROUP BY MONTH(r.rental_date), i.store_id) x
GROUP BY x.rental_month, x.store_id

我在任何地方都使用别名是明确的，你可能会省略一些。希望这有助于...

编辑:肮脏的技巧:

SELECT x.rental_month, x.store_id, MAX(x.rental_count) FROM (
SELECT MONTH(r.rental_date) AS rental_month, i.store_id AS store_id, COUNT(*) AS rental_count
FROM rental r LEFT JOIN inventory i ON r.inventory_id = i.inventory_id
WHERE YEAR(r.rental_date) = 2005
GROUP BY MONTH(r.rental_date), i.store_id
ORDER BY MONTH(r.rental_date) ASC, COUNT(*) DESC) x
GROUP BY x.rental_month

引用: http://kristiannielsen.livejournal.com/6745.html

但这会让您满意吗，因为您已经有了一个有效的查询...