获取带有统计信息的Rails模型

如果您正确声明 associations ,简单如下:

Product.find(:all, :include => :sales).each do |product|
  puts product.name + " (#{product.sales.size})"
end

编辑:
关联是丰富的集合,如果需要应用条件,可以对其进行搜索:

@products = Product.find(:all, :include => :sales)
@products.each do |product|
  # Find product sales only from yesterday
  yesterdays_sales = product.sales.find(:all, :conditions => { :created_at => Date.yesterday }
end

这不是一个答案,而是这个问题的三个答案的比较。我已经给出了答案。在工作中有一些困惑 :joins 和 :include 在里面 ActiveRecord.find . 所以我花了一些时间分析了三个解决方案的SQL日志。

方法1:使用 count

  #To return the sales count for each product
  ps_count_hash = Product.count(:joins => [:sales], :group => "products.id") # sql 1.1

  # To print the product id and sales count
  ps_count_hash.each do | product_id, sales_count|
   p "Product#{product_id} - (#{sales_count})"
  end

  # To print the product details and sales count
  # get all the products associated 
  Product.find_all_by_id(ps_count_hash.keys).each |product|  #sql 1.2
   p "Product[id = #{product.id}, name = #{product.name}] - (#{ps_count_hash[product.id]})"
  end

join

  Product.find(:all, :joins=>[:sales]).each do |product| #sql 2.1
   p "Product[id = #{product.id}, name = #{product.name}] - (#{product.sales.size})" # sql 2.2 - 2.(2+N)
  end

方法3:通过 include

  Product.find(:all, :include=>[:sales]).each do |product| #sql 3.1 and 3.2
   p "Product[id = #{product.id}, name = #{product.name}] - (#{product.sales.size})" 
  end

现在让我们看看这三种方法生成的SQL语句

方法1-2的SQL语句

SELECT count(*) AS count_all, products.id AS products_id FROM `products` INNER JOIN `sales` ON sales.product_id = products.id GROUP BY products.id 
SELECT `products`.* FROM `products` WHERE (`products`.`id` IN (1,2))

方法2-2的SQL语句

SELECT * FROM `products` 
SELECT `sales`.* FROM `sales` WHERE (`sales`.product_id IN (1,2)) 

方法3-n+1的SQL语句

SELECT `products`.* FROM `products` INNER JOIN `sales` ON sales.product_id = products.id 
SELECT * FROM `sales` WHERE (`sales`.product_id = 1) 
SELECT * FROM `sales` WHERE (`sales`.product_id = 2) 

按销售额计算产品的最佳方法(不含销售明细):

按销售额计算产品的最佳方法(含销售明细): 方法2

方法3有n+1问题。所以这是不争的。

如果你最感兴趣的是计算每种产品的销售额,你应该选择 counter_cache 特征。这将确保您始终从 products 表而不是JIT计数计算。

# products table should have a column called sales_count
class Product < ActiveRecord:Base
  has_many :sales
end

class Sales < ActiveRecord:Base
  belongs_to :product, :counter_cache => true
end

轨道将负责增加/减少 sales_count 创建/删除销售时的列。 现在,您可以通过执行以下操作来获取销售计数:

Product.first.sales_count

有条件地计算 sales 在单人间 product 对象执行以下操作:

Product.first.sales.count(:conditions => ["amount > ? ", 200])

有条件地计算一批 产品 s执行以下操作:

#returns a ordered hash
product_sales_count = Product.count(:joins => [:sales], :conditions => 
            ["amount > ? ", 200], :group =>"products.id").each do |product_id, sales_count|
   p "Product #{product_id} = #{sales_count}"
end

# If you need the product objects get all of them in one call and match the count
products = Product.find(:all, :joins => [:sales], :conditions => 
             ["amount > ? ", 200], :group =>"products.id").each do |product|

   p "Product #{product.name} = #{product_sales_count[product.id]}"

end

这种方法将节省大量到数据库的访问。

如果您不介意将数据库中的相关销售对象引入应用程序,那么 like this 应该做的工作是:

但要注意它。

# Preload all the Sales for our Products, sow we won't end up with `N+1` issue.
selected_products = Product.all :joins => :sales, :conditions => {'orders.successful' => true}

# Then use it like a normal association:
selected_products.each { |p| puts p.sales.size }

当做,
Dmitriy。