Question

执行join

后，我得到了这两个哈希数组

数组1

[#<State id: 1, name: "Alabama">, #<State id: 1, name: "Alabama">, #<State id: 1, name: "Alabama">, #<State id: 1, name: "Alabama">, #<State id: 2, name: "Alaska">, #<State id: 2, name: "Alaska">, #<State id: 4, name: "Arkansas">, #<State id: 4, name: "Arkansas">, #<State id: 4, name: "Arkansas">, #<State id: 6, name: "Colorado">, #<State id: 6, name: "Colorado">, #<State id: 6, name: "Colorado">, #<State id: 11, name: "Georgia">, #<State id: 14, name: "Illinois">, #<State id: 18, name: "Kentucky">, #<State id: 18, name: "Kentucky">, #<State id: 22, name: "Massachusetts">, #<State id: 48, name: "Washington">]

阵列2

[#<City id: 1, name: "Abbeville", state_id: 1>, #<City id: 1, name: "Abbeville", state_id: 1>, #<City id: 1, name: "Abbeville", state_id: 1>, #<City id: 4543, name: "Abingdon", state_id: 14>, #<City id: 8282, name: "Accord", state_id: 22>, #<City id: 3808, name: "Acworth", state_id: 11>, #<City id: 6855, name: "Adairville", state_id: 18>, #<City id: 6855, name: "Adairville", state_id: 18>, #<City id: 18895, name: "Adams County", state_id: 6>, #<City id: 4, name: "Addison", state_id: 1>, #<City id: 4, name: "Addison", state_id: 1>, #<City id: 17510, name: "Addy", state_id: 48>, #<City id: 1054, name: "Adona", state_id: 4>, #<City id: 1054, name: "Adona", state_id: 4>, #<City id: 577, name: "Akiachak", state_id: 2>, #<City id: 1056, name: "Alicia", state_id: 4>, #<City id: 583, name: "Ambler", state_id: 2>, #<City id: 2783, name: "Aspen", state_id: 6>]

我想根据每个数组中的state_id的值

从上面的两个数组中创建第三个数组例如，在这种情况下

[＃，....等等

为你的帮助我使用连接查询的前两个哈希数组

@states = State.joins("INNER JOIN property_of_interests ON property_of_interests.state_id = states.id").where(:property_of_interests => {:user_id => current_user.id})
@cities = City.joins("INNER JOIN property_of_interests ON property_of_interests.city_id = cities.id").where(:property_of_interests => {:user_id => current_user.id})

我可以处理查询本身以获得所需的输出吗？

我试过像

这样的东西

`@states.select("@states.name,@cities.name").joins("INNER JOIN @cities ON @cities.state_id = @states.id")`

但它不起作用。

更多信息

状态 id，name

城市 id，name，state_id

property_of_interests id，user_id，state_id，state_name

州有城市
城市属于州
州属于property_of_interests
城市属于property_of_interests
property_of_interests有城市
property_of_interests有州

所需的输出，如

State Name City Name
Alabama    Abbeville
Alabama    Abbeville
Alabama    Abbeville
....

Answer 1

您可以这样做：

City.all.each do |city|
  puts "#{city.state.name} #{city.name}"
end

或者，作为一个数组：

arr = City.all.map { |c| [c.state.name, c.name] }

或者作为哈希数组：

arr = City.all.map { |c| {state: c.state.name, city: c.name} }

或者实际回答这个问题，因为你想从properties_of_interest表开始：

PropertyOfInterest.all.each do |prop|
  prop.state.cities.each do |city|
    puts prop.state.name, city.name
  end
end

根据哈希内容查找两个哈希数组的交集

1 个答案: