Mongoid：选择适合多个选项的嵌入对象

Question

我有这个结构

class House
  include Mongoid::Document
  embeds_many :inhabitants
end

class Inhabitant
  include Mongoid::Document
  embedded_in :house
  field :name
  field :gender
  field :age
end

我可以得到所有女性居住的房屋：

houses = House.where("inhabitants.gender" => "female")

但我如何才能获得50岁以下女性居住的所有房屋？如何为嵌入对象指定多个条件？

Answer 1

要将多个条件应用于数组中的每个条目，您应该使用$elemMatch运算符。我不熟悉Mongoid，但是这里的MongoDB shell语法被修改为使用$elemMatch的查询：

> db.house.find({inhabitants: {$elemMatch: {gender: "female", age: {$lt: 50}}}})

Answer 2

试试这个：

houses = House.where("inhabitants.gender" => "female", "inhabitants.age" => {"$lt" => 50})

合并条件： MongoDB查询：

db.houses.find({'inhabitants.age' : {$in: {$lt: 50}}})

Mongoid：

houses = House.where('inhabitants.age' => {'$in' => {'$lt' => 50}})