你如何优化这个半复杂的mysql查询？

Question

问候伙伴们！

这是db模式（简化为相关领域），以帮助说明我的难题：

• 网上论坛有items，items有item_options。
• 相同的item可以出现在多个categories。
中
• categories，items和item_options都可以是有效或无效的（BOOL）。

类别看起来像这样（请注意parent_id嵌套，其中Fruit Seeds位于种子内部）：

id   parent_id   name              active
 1           0   Seeds                  1
 2           1   Vegetable Seeds        1
 3           1   Fruit Seeds            0
 4           0   Plants                 1
 5           4   Vegetable Plants       1
 6           4   Fruit Plants           1

我想要的是所有活动类别（id，parent_id和name）的快速列表，以及包含活动item_options的每个类别的活动项目的计数。

查询结果如下所示：

id   parent_id   name              item_count
 1           0   Seeds                      0
 2           1   Vegetable Seeds           52
 4           0   Plants                     0
 5           4   Vegetable Plants         103
 6           4   Fruit Plants              79

此查询有效，但需要~430ms ：

    SELECT c.`id`, c.`parent_id`, c.`name`,
        (SELECT COUNT(*)
            FROM `item_categories` AS ic
            LEFT JOIN `items` AS i
                ON (i.`id` = ic.`item_id`)
            LEFT JOIN `item_options` AS io
                ON (i.`id` = io.`item_id`)
            WHERE c.`id` = ic.`category_id`
                AND i.`active` = 1
                AND io.`active` = 1
            ) AS `item_count`
    FROM `categories` AS c
    WHERE c.`active` = 1;

下一个查询只需要约55毫秒，但无法包含顶级类别（parent_id = 0）：

    SELECT c.`id`, c.`parent_id`, c.`name`,
        COUNT(ic.`item_id`) AS `item_count`
    FROM `categories` AS c
    LEFT JOIN `item_categories` AS ic
        ON (c.`id` = ic.`category_id`)
    LEFT JOIN `items` AS i
        ON (i.`id` = ic.`item_id`)
    LEFT JOIN `item_options` AS io
        ON (i.`id` = io.`item_id`)
    WHERE c.`active` = 1
      AND i.`active` = 1
      AND io.`active` = 1
    GROUP BY c.`id`;

任何人都会看到如何加快第一个查询，或修复第二个查询？

Answer 1

哦！这似乎是这样做的......

(
    SELECT c.`id`, c.`parent_id`, c.`name`,
        COUNT(ic.`item_id`) AS `item_count`
    FROM `categories` AS c
    LEFT JOIN `item_categories` AS ic
        ON (c.`id` = ic.`category_id`)
    LEFT JOIN `items` AS i
        ON (i.`id` = ic.`item_id`)
    LEFT JOIN `item_options` AS io
        ON (i.`id` = io.`item_id`)
    WHERE c.`active` = 1
      AND i.`active` = 1
      AND io.`active` = 1
    GROUP BY c.`id`
)
UNION
(
    SELECT c.`id`, c.`parent_id`, c.`name`, 0
    FROM `categories` AS c
    WHERE c.`parent_id` = 0
      AND c.`active` = 1
)
ORDER BY `parent_id`, `NAME`;

有人看到更好的方法吗？ ：d