Question

我不是mysql专家，遇到了以下情况。

我有以下3个表，即article，tag和articles_tags，试图以尽可能少的方式描述表的定义。

article: id
tag: id
articles_tags: article_id, tag_id

现在我需要执行以下操作

获取所有标签
然后是第1步中每个标签的文章。
然后再次从步骤2获取每个文章的所有标签，然后根据它们的出现将它们分组，并以json的形式获得最终结果。

例如名为whatsapp的标签，结果下面将一行显示。

whatsApp， WhatsAppDisplay，[{“slug”:“messenger”,“displayName”:“Messenger”,“frequency”:159},{“slug”:“apps”,“displayName”:“Apps”,“frequency”:148},{“slug”:“chat”,“displayName”:“Chat”,“frequency”:109},{“slug”:“smartphone”,“displayName”:“Smartphone”,“frequency”:79},{“slug”:“iphone”,“displayName”:“iPhone”,“frequency”:54},{“slug”:“app”,“displayName”:“App”,“frequency”:53},{“slug”:“android”,“displayName”:“Android”,“frequency”:27},{“slug”:“handy”,“displayName”:“Handy”,“frequency”:26},{“slug”:“sprueche”,“displayName”:“Spr\u00fcche”,“frequency”:25},{“slug”:“ios”,“displayName”:“iOS”,“frequency”:25},{“slug”:“facebook”,“displayName”:“Facebook”,“frequency”:24},{“slug”:“nachrichten”,“displayName”:“Nachrichten”,“frequency”:22},{“slug”:“bilder”,“displayName”:“Bilder”,“frequency”:20},{“slug”:“android-smartphone”,“displayName”:“Android-Smartphone”,“frequency”:16},{“slug”:“kontakte”,“displayName”:“Kontakte”,“frequency”:15}]

为此，我编写了以下sql查询。

    SELECT t3.slug as testSlug, t3.display_name AS displayName,
    (
        SELECT CONCAT('[', second_json, ']') AS final_json FROM
            (
                SELECT GROUP_CONCAT('{', first_json, '}' SEPARATOR ',') AS second_json FROM
                    (
                        SELECT
                            CONCAT
                                (
                                '"slug":'   , '"', t2.slug   , '"', ',' 
                                '"displayName":', '"', t2.display_name, '"', ','
                                '"frequency":'  , count(t2.slug)
                                ) AS first_json FROM tipps.tag t2 
                                INNER JOIN tipps.articles_tags ats2 ON ats2.tag_id = t2.id
                                INNER JOIN tipps.article a2 ON a2.id = ats2.article_id
                                WHERE a2.id IN (
                                    SELECT a.id FROM tipps.tag t
                                    INNER JOIN tipps.articles_tags ats ON ats.tag_id = t.id
                                    INNER JOIN tipps.article a ON a.id = ats.article_id
                                    WHERE t.slug = 'whatsapp'
                                )
                                AND t2.slug != 'whatsapp'
                                GROUP BY t2.slug
                                ORDER BY count(t2.slug) DESC
                                LIMIT 15
                    ) AS third_json
            ) AS fourth_json
    ) AS relatedTags
FROM tipps.tag t3
LIMIT 1;
FROM tipps.tag t3 LIMIT 1;

对于那些真正难以理解查询的人。请在下面找到简单版本。

    /* step 2 find all tags belongs to each article */
    SELECT 
    t2.slug, 
    count(t2.slug) AS frequency 
FROM tipps.tag t2 
    INNER JOIN tipps.articles_tags ats2 ON ats2.tag_id = t2.id
    INNER JOIN tipps.article a2 ON a2.id = ats2.article_id
WHERE a2.id IN (

        /* step one find all articles belongs to a tag */
        SELECT a.id 
        FROM tipps.tag t
            INNER JOIN tipps.articles_tags ats ON ats.tag_id = t.id
            INNER JOIN tipps.article a ON a.id = ats.article_id
        WHERE t.slug = 'whatsapp'


    )

/* group them based on their occurances and sort based on occurances */
GROUP BY t2.slug
ORDER BY frequency DESC
LIMIT 15;

可以看出，我使用whatsapp作为常量，现在可以传递t3.tipps而不是常量。这样我就可以在一个查询中选择所有结果。还是对此有更好的方法。任何帮助或线索都会很棒。

Answer 1

我们要获取标记的所有文章以及这些文章的所有标记。这是获取所需所有数据的原始方法：

select article.id as article_id, tag.id as first_tag_id, article_tags2.tag_id as second_tag_id
from tag
join article_tags article_tags1
on tag.id = article_tags1.tag_id
join article
on article_tags1.article_id = article.id
join article_tags article_tags2
on article.id = article_tags2.article_id

现在我们有了所有标签，每个标签的所有文章以及每个文章的所有标签，我们可以将结果分组：

select article.id as article_id, tag.id as first_tag_id, article_tags2.tag_id as second_tag_id
from tag
join article_tags article_tags1
on tag.id = article_tags1.tag_id
join article
on article_tags1.article_id = article.id
join article_tags article_tags2
on article.id = article_tags2.article_id
group by tag.id, article.id

从这里开始，您可以对适当的组进行group_concat和其他计算。

mysql将选择结果传递给另一个查询

1 个答案: