Ruby on rails 3 Rails查询返回属于任何城市的用户&；不属于任何城市

我有两个表之间的

多对多关联

：针对前用户和城市

users
id  name
1   Bob
2   Jon
3   Tom
4   Gary
5   Hary

cities
id     name 
1      London
2      New-york
3      Delhi

users_cities
id   user_id   city_id
1    1         2
2    2         1
3    3         1
4    3         2
5    4         3

我想要两个sql查询

接受城市id数组并返回属于该城市的所有用户的查询。 例如，当city_id:[1，2]时，结果应为 O/P应为

   id  name
    1   Bob
    2   Jon
    3   Tom

接受城市id数组并返回不属于这些城市的所有用户的查询。 例如，当city_id:[1，2]时，结果应为 O/P应为

    id  name
    4   Gary
    5   Hary

注意：-我正在使用

user.rb

has_and_belongs_to_many :cities

has_and_belongs_to_many :users

city.rb

has_and_belongs_to_many :cities

has_and_belongs_to_many :users

---

基本上，您需要两个方法/作用域

class User < ActiveRecord::Base
  has_and_belongs_to_many :cities

  scope :by_cities, ->(city_ids) {includes(:cities).where(cities: {id: city_ids}).distinct}

  # There are several ways to do that
  # 1. That will return all not associated records but that we won't need
  # In this scenario, You need OR condition like city_ids OR NULL 
  # This will return => ["Hary"] 
  scope :not_by_cities, -> {includes(:cities).where(cities: {id: nil})}

  # 2. Need to create a scope in City model
  # scope :city_ids, -> all.pluck(:id) 
  # This will return => ["Gary", "Hary"] 
  scope :not_by_cities, -> {includes(:cities).where(cities: {id: [City.city_ids - city_ids, nil]})}

  # 3. If you are on Rails 5, It is much more easier
  # This will return => ["Gary", "Hary"] 
  scope :not_by_cities, -> {includes(:cities).where.not(cities: {id: city_ids}).where(cities: {id: nil})} 
end

如果您是Rails4.x，仍然需要一些简单的解决方案。有人在那里吗

希望这对你有帮助