Sequelize 通过手动定义的关联查找 table

Question

我知道描述了一个更简单的案例 here:

不幸的是，我的情况比那复杂一点。我有一个用户模型，其中 belongsToMany 个部门（反过来 belongsToMany 个用户），但是通过 userDepartment，一个手动定义的连接 table 来实现。 我的目标是获取属于给定部门的所有用户。首先让我们看一下models/user.js：

var user = sequelize.define("user", {
    id: {
        type: DataTypes.INTEGER,
        field: 'emplId',
        primaryKey: true,
        autoIncrement: false
    },
    firstname: {
        type: DataTypes.STRING,
        field: 'firstname_preferred',
        defaultValue: '',
        allowNull: false
    }
    ...
    ...
    ...

    associate: function(models) {
            user.belongsToMany(models.department, {
                foreignKey: "emplId",
                through: 'userDepartment'
                });
            })
    }
    ...
    return user;

现在，看看 models/department.js：

var department = sequelize.define("department", {
    id: {
        type: DataTypes.INTEGER,
        field: 'departmentId',
        primaryKey: true,
        autoIncrement: true
    },
    ...
    classMethods: {
        associate: function(models) {

            department.belongsToMany(models.user, {
                foreignKey: "departmentId",
                through: 'userDepartment',
                onDelete: 'cascade'
            });
        }

    ...
return department;

最后在 models/userDepartment.js：

var userDepartment = sequelize.define("userDepartment", {
    title: {
        type: DataTypes.STRING,
        field: 'title',
        allowNull: false,
        defaultValue: ''
    }
}, {
    tableName: 'user_departments'
});

return userDepartment;

到目前为止一切顺利。然而，这个查询：

models.user.findAll({
    where: {'departments.id': req.params.id},
    include: [{model: models.department, as: models.department.tableName}]

})

失败并出现以下错误：

SequelizeDatabaseError: ER_BAD_FIELD_ERROR: Unknown column 'user.departments.id' in 'where clause'

尝试包含 userDepartment 模型会导致：

Error: userDepartment (user_departments) is not associated to user!

简而言之：我有两个具有 M:M 关系的 Sequelize 模型。它们通过手动定义的联接 table 关联（它为每个唯一关系添加职位，即用户 A 是部门 B 中的 "Manager"）。尝试按部门查找用户失败，出现严重的 table 名称错误。

续集版本“^2.0.5”

Answer 1

花了几个小时，但我找到了我的解决方案：

models.department.find({
    where: {id:req.params.id},
    include: [models.user]

问题是 Sequelize 不允许您 "go out of scope" 因为它以 model_name 开始每个 where 子句。因此，例如，当部门 table 仅作为 departments.id 加入时，where 子句试图比较 user.departments.id。由于我们正在查询部门的值（ID），因此查询单个部门和 return 他们的关联用户是最有效的。

Answer 2

我有一个类似的问题，但在我的情况下我无法切换表格。

我必须使用：sequelize.literal 函数。

在您的情况下，它将如下所示：

models.user.findAll({
    where: sequelize.literal("departments.id = " + req.params.id),
    include: [{model: models.department, as: models.department.tableName}]
})

我不喜欢它，但它有效。

Answer 3

对于仍在为此寻找答案的任何人，我在 Github 上找到了一个答案。

你只需这样做：

where: {
   '$Table.column$' : value
 }

Answer 4

如果您有 class 实例，您也可以使用自动生成的 instance.getOthers() 方法

这可能意味着一个额外的查询。但是如果实例已经在手边，这是最方便的语法。

假设“用户喜欢给定分数的 post”情况，我们可以得到用户喜欢的所有 post：

const user0 = await User.create({name: 'user0'})
const user0Likes = await user0.getPosts({order: [['body', 'ASC']]})
assert(user0Likes[0].body === 'post0');
assert(user0Likes[0].UserLikesPost.score === 1);
assert(user0Likes.length === 1);

完整的可运行示例：

main.js

const assert = require('assert')
const { DataTypes, Op, Sequelize } = require('sequelize')
const common = require('./common')
const sequelize = common.sequelize(__filename, process.argv[2], { define: { timestamps: false } })
;(async () => {

// Create the tables.
const User = sequelize.define('User', {
  name: { type: DataTypes.STRING },
});
const Post = sequelize.define('Post', {
  body: { type: DataTypes.STRING },
});
const UserLikesPost = sequelize.define('UserLikesPost', {
  UserId: {
    type: DataTypes.INTEGER,
    references: {
      model: User,
      key: 'id'
    }
  },
  PostId: {
    type: DataTypes.INTEGER,
    references: {
      model: Post,
      key: 'id'
    }
  },
  score: {
    type: DataTypes.INTEGER,
  },
});
User.belongsToMany(Post, {through: UserLikesPost});
Post.belongsToMany(User, {through: UserLikesPost});
await sequelize.sync({force: true});

// Create some users and likes.

const user0 = await User.create({name: 'user0'})
const user1 = await User.create({name: 'user1'})
const user2 = await User.create({name: 'user2'})

const post0 = await Post.create({body: 'post0'});
const post1 = await Post.create({body: 'post1'});
const post2 = await Post.create({body: 'post2'});

// Autogenerated add* methods

// Make some useres like some posts.
await user0.addPost(post0, {through: {score: 1}})
await user1.addPost(post1, {through: {score: 2}})
await user1.addPost(post2, {through: {score: 3}})

// Find what user0 likes.
const user0Likes = await user0.getPosts({order: [['body', 'ASC']]})
assert(user0Likes[0].body === 'post0');
assert(user0Likes[0].UserLikesPost.score === 1);
assert(user0Likes.length === 1);

// Find what user1 likes.
const user1Likes = await user1.getPosts({order: [['body', 'ASC']]})
assert(user1Likes[0].body === 'post1');
assert(user1Likes[0].UserLikesPost.score === 2);
assert(user1Likes[1].body === 'post2');
assert(user1Likes[1].UserLikesPost.score === 3);
assert(user1Likes.length === 2);

// Where on the custom through table column.
// Find posts that user1 likes which have score greater than 2.
// 
{
  const rows = await Post.findAll({
    include: [
      {
        model: User,
        where: {id: user1.id},
        through: {
          where: {score: { [Op.gt]: 2 }},
        },
      },
    ],
  })
  assert.strictEqual(rows[0].body, 'post2');
  // TODO how to get the score here as well?
  //assert.strictEqual(rows[0].UserLikesPost.score, 3);
  assert.strictEqual(rows.length, 1);
}

})().finally(() => { return sequelize.close() });

common.js

const path = require('path');

const { Sequelize } = require('sequelize');

function sequelize(filename, dialect, opts) {
  if (dialect === undefined) {
    dialect = 'l'
  }
  if (dialect === 'l') {
    return new Sequelize(Object.assign({
      dialect: 'sqlite',
      storage: path.parse(filename).name + '.sqlite'
    }, opts));
  } else if (dialect === 'p') {
    return new Sequelize('tmp', undefined, undefined, Object.assign({
      dialect: 'postgres',
      host: '/var/run/postgresql',
    }, opts));
  } else {
    throw new Error('Unknown dialect')
  }
}
exports.sequelize = sequelize

package.json

{
  "name": "tmp",
  "private": true,
  "version": "1.0.0",
  "dependencies": {
    "pg": "8.5.1",
    "pg-hstore": "2.3.3",
    "sequelize": "6.5.1",
    "sqlite3": "5.0.2"
  }
}

在 PostgreSQL 13.4、Ubuntu 21.04.

上测试

Sequelize 通过手动定义的关联查找 table

Sequelize find by association through manually-defined join table

node.js

sequelize.js