Sequelize 会在 where 子句中自动添加子查询。有没有办法让它跳过添加 where 子句

Question

我有一个使用 INNER JOINS 的 Sequelize 查询。问题是 sequelize 在内部添加了另一个 where 子句和子表上的子查询。那正在消耗查询性能。下面是我的代码示例和原始查询输出。

有没有办法让 sequelize 跳过添加这个 where 子句？

续集版本：6.x

PostModel.findAll({
   where: {
     id: 1,
   },
   include: [
     {
        model: CommentsModel,
        required: true,
     }
   ]
})

查询构建如下 SQL 查询。

SELECT "post".*
FROM (SELECT "post"."*"
      FROM "posts" AS "post"
      WHERE "post"."id" = 2
        AND (SELECT "post_id"
             FROM "comments" AS "c"
             WHERE "comments"."post_id" = "post"."id" AND ("c"."text_search" @@ 'who:*')) IS NOT NULL
      ORDER BY "post"."id" DESC
      LIMIT 50 OFFSET 0) AS "post"
         LEFT OUTER JOIN "post_tags" AS "tags" ON "post"."id" = "tags"."post_id"
         LEFT OUTER JOIN "tag" AS "tags->tag" ON "tags"."tag_id" = "tags->tag"."id"
         INNER JOIN "comments" AS "c" ON "post"."id" = "c"."post_id" AND ("c"."text_search" @@ 'who:*')
ORDER BY "post"."id" DESC;

如您所见，WHERE 子句新增了一个

(SELECT "post_id"
             FROM "comments" AS "c"
             WHERE "comments"."post_id" = "post"."id" AND ("c"."text_search" @@ 'who:*'))

这基本上是在扼杀查询的性能。

Answer 1

经过大量研究，我找到了解决方案。

我们需要在关联中添加subQuery: false。

PostModel.findAll({
   where: {
     id: 1,
   },
   include: [
     {
        subQuery: false,
        model: CommentsModel,
        required: true,
     }
   ]
})

查询输出：

SELECT "post".*
FROM (SELECT "post"."*"
      FROM "posts" AS "post"
      WHERE "post"."id" = 2 
      ORDER BY "post"."id" DESC
      LIMIT 50 OFFSET 0) AS "post"
         LEFT OUTER JOIN "post_tags" AS "tags" ON "post"."id" = "tags"."post_id"
         LEFT OUTER JOIN "tag" AS "tags->tag" ON "tags"."tag_id" = "tags->tag"."id"
         INNER JOIN "comments" AS "c" ON "post"."id" = "c"."post_id" AND ("c"."text_search" @@ 'who:*')
ORDER BY "post"."id" DESC;