javascript - Sequelize JS - 在连接中选择 On 子句

Question

我想知道如何在连接中选择 ON 子句。我有两个表，packages 和 users。 package 表有两个字段/列(owner_id、helper_id)，它们是同一表 users 的外键。我想要一个用于加入 packages.owner_id = users.id 和 packages.helper_id = users.id 的位置。这是我所拥有的:

models.Packages.findAll({
    where: (!package_id ? { owner_id } : { owner_id, id: { [Op.lt]: package_id } }),
    include: [{
      model: models.Users,
      attributes: { exclude: ['password'] },
      where: ['helper_id = id']
    }],
    limit: 5,
    order: [["id","DESC"]]
  })

还有另一个

models.Packages.findAll({
    where: (!package_id ? { owner_id } : { owner_id, id: { [Op.lt]: package_id } }),
    include: [{
      model: models.Users,
      attributes: { exclude: ['password'] },
      where: ['owner_id = id']
    }],
    limit: 5,
    order: [["id","DESC"]]
  })

但我收到此错误:

Unhandled rejection Error: Support for literal replacements in the `where` object has been removed

请帮忙？

Answer 1

我喜欢 Shahar Hadas 的回答，但我确实找到了如何解决我想做的事情。

基本上，我必须定义具有特定别名的关系/关联(注意 as 属性):

models.Users.hasMany(models.Packages, { foreignKey: 'owner_id', sourceKey: 'id' });
models.Users.hasMany(models.Packages, { foreignKey: 'helper_id', sourceKey: 'id' });
models.Packages.belongsTo(models.Users, { as: 'owner', foreignKey: 'owner_id', targetKey: 'id' });
models.Packages.belongsTo(models.Users, { as: 'helper', foreignKey: 'helper_id', targetKey: 'id' });

然后当我查询时，我可以选择我想要的别名:

models.Packages.findAll({
    where: (!package_id ? { owner_id } : { owner_id, id: { [Op.lt]: package_id } }),
    include: [{
      model: models.Users,
      as: 'owner',
      attributes: { exclude: ['password'] }
    }],
    limit: 5,
    order: [["id","DESC"]]
  })

或

models.Packages.findAll({
    where: (!package_id ? { owner_id } : { owner_id, id: { [Op.lt]: package_id } }),
    include: [{
      model: models.Users,
      as: 'helper',
      attributes: { exclude: ['password'] }
    }],
    limit: 5,
    order: [["id","DESC"]]
  })