合并来自 2 个 SQL 结果的数组

Question

所以我在 SQL 中有一个标签表设置

列表、列表标签、标签

每个列表可以有多个标签，标签有流派和制片人两种。现在，如果我们正在寻找所有带有标签“动作”的列表，这些是我正在遵循的步骤

SELECT GROUP_CONCAT(mini_lists_tags.list_id) AS list_ids
FROM (`mini_tags`)
LEFT JOIN `mini_lists_tags` ON `mini_lists_tags`.`tag_id` = `mini_tags`.`tag_id`
WHERE `mini_tags`.`tag_slug` = 'action'  

这将为列表的 id 返回一个数组 1,2。

SELECT *
FROM (`mini_lists_anime`)
JOIN `mini_lists` ON `mini_lists`.`list_id` = `mini_lists_anime`.`list_id`
WHERE `mini_lists`.`list_id` IN ('1', '2') 
AND `mini_lists`.`list_state` = 'active'

获取数组中的所有列表示例：

Array
(
    [0] => stdClass Object
        (
            [list_id] => 1
            [list_episodes] => 13
            [list_duration] => 24
            [list_aired] => 1238623200
            [list_age_rate] => PG-13 - Teens 13 or older
            [user_id] => 1
            [list_mal] => 5342
            [list_category] => Anime
            [list_type] => TV
            [list_status] => Completed
            [list_title] => Asura Cryin'
            [list_alt_titles] => アスラクライン
            [list_thumb] => 17071
            [list_likes] => 0
            [list_date] => 1300609723
            [list_update] => 0
            [list_state] => active
            [list_info] => 
        )

    [1] => stdClass Object
        (
            [list_id] => 2
            [list_episodes] => 26
            [list_duration] => 23
            [list_aired] => 1238623200
            [list_age_rate] => PG-13 - Teens 13 or older
            [user_id] => 1
            [list_mal] => 329
            [list_category] => Anime
            [list_type] => TV
            [list_status] => Completed
            [list_title] => Planetes
            [list_alt_titles] => プラネテス
            [list_thumb] => 4822
            [list_likes] => 0
            [list_date] => 1300609723
            [list_update] => 0
            [list_state] => active
            [list_info] => 
        )

)

然后我们得到标签

SELECT `mini_lists_tags`.`list_id`, `mini_tags`.`tag_type`, GROUP_CONCAT(mini_tags.tag_name) AS tag_names
FROM (`mini_lists_tags`)
INNER JOIN `mini_tags` ON `mini_tags`.`tag_id` = `mini_lists_tags`.`tag_id`
WHERE `mini_lists_tags`.`list_id` IN ('1', '2') 
GROUP BY `mini_lists_tags`.`list_id`, `mini_tags`.`tag_type`  

获取数组中的所有标签示例：

Array
(
    [0] => stdClass Object
        (
            [list_id] => 1
            [tag_type] => Genre
            [tag_names] => Supernatural,Action,Mecha
        )

    [1] => stdClass Object
        (
            [list_id] => 1
            [tag_type] => Producers
            [tag_names] => Seven Arcs
        )

    [2] => stdClass Object
        (
            [list_id] => 2
            [tag_type] => Genre
            [tag_names] => Romance,Action,Sci-fi,Comedy,Slice of Life,Drama,Space
        )

    [3] => stdClass Object
        (
            [list_id] => 2
            [tag_type] => Producers
            [tag_names] => Sunrise,Bandai Entertainment,Bandai Visual,Bang Zoom! Entertainment
        )

)

现在的问题是我需要将它们合并到 list_id 上，这样它就会为每个返回类似这样的内容。

    stdClass Object
    (
        [list_id] => 1
        [list_episodes] => 13
        [list_duration] => 24
        [list_aired] => 1238623200
        [list_age_rate] => PG-13 - Teens 13 or older
        [user_id] => 1
        [list_mal] => 5342
        [list_category] => Anime
        [list_type] => TV
        [list_status] => Completed
        [list_title] => Asura Cryin'
        [list_alt_titles] => アスラクライン
        [list_thumb] => 17071
        [list_likes] => 0
        [list_date] => 1300609723
        [list_update] => 0
        [list_state] => active
        [list_info] => 
        [list_tags] => Array
            (
                [0] => stdClass Object
                    (
                        [tag_type] => Genre
                        [tag_names] => Mecha,Action,Supernatural
                    )

                [1] => stdClass Object
                    (
                        [tag_type] => Producers
                        [tag_names] => Seven Arcs
                    )

            )

)

感谢任何建议，我真的迷路了。如果有比这更好的解决方案，我会全力以赴。

Answer 1

您可以执行另一种类型的联接，该联接将多次返回与每个子对象合并的父项，如下所示：

Array(
   [0] => stdClass Object
    (
        [list_id] => 1
        [list_episodes] => 13
        [list_duration] => 24
        ...etc
        [tag_type] => Genre
        [tag_names] => Supernatural,Action,Mecha
        ...etc
    )
   [1] => stdClass Object
    (
        [list_id] => 1
        [list_episodes] => 13
        [list_duration] => 24
        ...etc
        [tag_type] => Producers
        [tag_names] => Seven Arcs
        ...etc
    )
   [2] => stdClass Object
    (
        [list_id] => 2
        [list_episodes] => 26
        [list_duration] => 23
        ...etc
        [tag_type] => Genre
        [tag_names] => Supernatural,Action,Mecha
        ...etc
    )
   [3] => stdClass Object
    (
        [list_id] => 2
        [list_episodes] => 26
        [list_duration] => 23
        ...etc
        [tag_type] => Producers
        [tag_names] => Seven Arcs
        ...etc
    )
)

然后，您需要遍历您的结果，将结果合并到他们的子/父关系中。这是因为 SQL 总是将行作为结果返回，而不是复杂的结构。

虽然处理起来更复杂，但它通常比为每个父对象进行循环 sql 查询（称为 n+1 查询）的过程密集度更低