对于选项2

Question

我在两个表之间有Many to Many Associations：对于Ex用户＆amp;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

查询接受city_id数组并返回属于该城市的所有用户。 对于Ex，当city_id：[1,2]时，结果应为 O / P应该是

   id  name
    1   Bob
    2   Jon
    3   Tom

查询接受city_id数组并返回不属于这些城市的所有用户。 对于Ex，当city_id：[1,2]时，结果应为 O / P应该是

    id  name
    4   Gary
    5   Hary

注意： - 我正在使用

user.rb

has_and_belongs_to_many :cities

city.rb

has_and_belongs_to_many :users

Answer 1

基本上你需要两个方法/范围

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

对于选项2

class City < ActiveRecord::Base
    has_and_belongs_to_many :cities
    scope :city_ids, -> {all.pluck(:id)}
 end

结果：

>> User.by_cities([1,2]).pluck(:name)
=> ["Bob", "Jon", "Tom"]

>> User.not_by_cities.pluck(:name)
=> ["Gary", "Hary"]

如果你是Rails 4.x并且仍然想要一些简单的解决方案。使用那里的任何人

希望这会对你有所帮助。

Rails查询返回用户属于任何城市＆amp;不属于任何城市

1 个答案:

对于选项2