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问候flowstackers同胞！

这是 db 模式（简化为相关字段）以帮助说明我的难题：

在此处输入图像描述

  • 网上商店有items，也items有item_options。
  • 同一个item可以出现在多个categories.
  • categories,items并且item_options都可以是活动的或非活动的 (BOOL)。

类别看起来像这样（注意 parent_id 嵌套，其中水果种子位于种子内部）：

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 和名称）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，但未能包含顶级类别（其中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`;

有人知道如何加快第一个查询或修复第二个查询吗？

4

1 回答 1

0

哦！这似乎做到了......

(
    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

于 2012-09-27T12:37:28.673 回答