跨多个表创建 Sequelize 查询、多个内部连接和 Wheres

Question

我创建了一个成功的 mysql 查询，该查询连接了 2 个彼此具有外键的表。除了让用户附加到响应之外，我在 Sequelize 中几乎完成了所有工作。我的查询成功了，但我不明白如何获取它。

这是我的 mysql 查询，它专门获取与响应“user_id”

具有相同 ID 的“用户”

SELECT *
FROM meetup_db.posts p
INNER JOIN meetup_db.responses r
INNER JOIN meetup_db.users u
WHERE p.id = r.post_id AND
r.user_id = u.id

这是我的 sequelize 查询，几乎可以工作，但不会在 user_id = id

的每个响应中返回用户

const dbPostData = await Posts.findOne({
    where: { id: req.params.id },
    include: [
        { model: Users },
        {
            model: Responses,
            include: {
                model: Users,
                where: { id: Responses.user_id }, //this is the line in question i need the user that has the same id as the Response.user_id
            }
        }
    ]
});

sequelize 正确输出除用户以外的所有内容，这是一个示例输出，您可以看到响应中列出了 Users，但用户仅显示为 [对象]

{
  id: 1,
  title: 'BBQ At My House!!',
  description: 'Hey gang! Im having a BBQ at my house! I hope all can attend!!',
  upvotes: 44,
  location: '2113 Main St. Austin, TX.',
  date_occuring: 2021-08-04T18:00:00.000Z,
  created_at: 2021-08-06T04:42:01.000Z,
  edited: false,
  user_id: 1,
  createdAt: 2021-08-06T04:42:01.000Z,
  updatedAt: 2021-08-06T04:42:01.000Z,
  User: {
    id: 1,
    username: 'Jennifer Wylan',
    email: 'jwylan@gmail.com',
    password: 'ewfchijwnsj',
    image_url: '/assets/fake-pfp/fakeperson1.png'
  },
  Responses: [
    {
      id: 4,
      response: 'Im in there like swimwear!',
      user_id: 4,
      post_id: 1,
      createdAt: 2021-08-06T04:42:01.000Z,
      updatedAt: 2021-08-06T04:42:01.000Z,
      User: [Object] //these lines right here need to look like the above User: {} Object
    },
    {
      id: 5,
      response: 'weeeeeeeeeeeeeeeeeeeeeeeeeeeeee',
      user_id: 1,
      post_id: 1,
      createdAt: 2021-08-06T04:42:01.000Z,
      updatedAt: 2021-08-06T04:42:01.000Z,
      User: [Object] //these lines right here need to look like the above User: {} Object
    }
  ],
}

Answer 1

我明白了。所以它已经起作用了。在我的终端中，响应的输出将用户密钥显示为 [Object] ...原来当它像 3 个表深时，终端只显示 [Object] 但它实际上已经在那里，你也不需要 where: 因为它已经在寻找外键

跨多个表创建 Sequelize 查询、多个内部连接和 Wheres

Create Sequelize Query, Multiple Innerjoins and Wheres across multiple tables

javascript

mysql

node.js

sequelize.js