Sql 在搜索查询中附加内部联接的替代方法

搜索帖子时，用户可以从要筛选的类别列表中进行选择，这将返回与所选所有类别相关的帖子

SELECT *
FROM posts
INNER JOIN category_joins as filter_categories_join
  ON filter_categories_join.categorizable_id = posts.id 
  AND filter_categories_join.categorizable_type = 'Post'
INNER JOIN categories as filter_categories 
  ON filter_categories.id = filter_categories_join.category_id
INNER JOIN category_joins as categories_join_1
  ON categories_join_1.categorizable_id = posts.id
  AND categories_join_1.categorizable_type = 'Post'
INNER JOIN categories as categories_1
  ON categories_1.id = categories_join_1.category_id
INNER JOIN category_joins as categories_join_2
  ON categories_join_2.categorizable_id = posts.id
  AND categories_join_2.categorizable_type = 'Post'
INNER JOIN categories as categories_2
  ON categories_2.id = categories_join_2.category_id
WHERE filter_categories.lft BETWEEN 14 AND 115
  AND categories_1.lft BETWEEN 133 AND 134
  AND categories_2.lft BETWEEN 137 AND 138

有没有办法避免为每个类别筛选器追加内部联接

更新：

此处对关联的描述：

• 帖子

  有许多

  类别

• 帖子

  与

  类别

  相关，通过

  类别加入

  可分类

• categories

  通过

  categories\u连接有许多
  帖子

---

不确定为什么要复制不必要的

连接。您的查询可以简化为

SELECT p.*
FROM posts p
INNER JOIN category_joins cj
  ON cj.categorizable_id = p.id 
  AND cj.categorizable_type = 'Post'

INNER JOIN categories c
  ON c.id = cj.category_id
AND c.lft BETWEEN 14 AND 115
    INNER JOIN categories c1
  ON c1.id = cj.category_id
  AND c1.lft BETWEEN 133 AND 134
INNER JOIN categories c2
  ON c2.id = cj.category_id
 AND c2.lft BETWEEN 137 AND 138;

你迫使我们猜测，请你提供示例数据并描述数据的行为结构。例如，我假设
posts
和
category\u连接之间的连接是
1:many
？
category\u join
和
filter\u categories
之间的连接是
1:1
？你需要提供更多的信息才能得到一个完整的答案。