通过具有额外条件的联接获取所有记录

Question

“邻里”属于“城市”。 “城市”有很多“社区”。

如何通过城市名称选择属于特定城市的所有社区？城市名称是存储在城市表中的属性。

Neighborhood.joins(:city)将使用此SQL获取所有社区：

SELECT "neighborhoods".* 
FROM "neighborhoods" 
INNER JOIN "cities" 
ON "cities"."id" = "neighborhoods"."city_id"

但是使用这个SQL执行查询的Rails方式是什么（唯一的区别是最后一行）？

SELECT "neighborhoods".* 
FROM "neighborhoods" 
INNER JOIN "cities" 
ON "cities"."id" = "neighborhoods"."city_id"
AND "cities"."name" = "New York"

Answer 1

只需使用cities.name：

应用where条件

Neighborhood.joins(:city).where("cities.name = ?", "New York")

Answer 2

执行此操作的'rails'方式是在模型中使用活动记录关联

class City < ActiveRecord::Base   
  has_many :neighborhoods
  ....
end

class Neighborhood < ActiveRecord::Base
  belongs_to :city
  ...
end

然后你可以打电话：

neighborhoods = City.find_by(name: "New York").neighborhoods

Answer 3

如果使用Rails模型，不完全是sql，但结果可能是：

city = City.where（：name =＆gt;“New York”）。首先 city.neighborhoods

Answer 4

您应该尝试使用此代码 -

    City.joins(:neighborhoods).select("neighborhoods.*").where("cities.name = 'New York'")

阐释 -

It will be 'City.joins(:neighbourhoods)' because City has_many neightbours and 'city_id' will work as foreign key for neighbourhood model.

select（“neighborhoods。*”）将从邻域表中选择所有列。

其中（“cities.name ='New York'”）是您的所有记录，其中“城市名称”等于“纽约”。