Ruby on Rails 变量在视图中显示 Active Record 数据而不是整数

Question

我希望帮助器中有两个值是根据 Active Record 数据计算的，它们显示在同一个视图中，“gross_income_upload”和“all_time_gross”。 总收入还可以，但没有显示所有时间的总整数，而是显示的是 Active Record 数组数据。

有一个辅助函数和一个 .erb 视图，代码如下：

应用程序助手（application_helper.rb）：

module ApplicationHelper

  def gross_income_upload
    if Sale.any?
      gross_income_upload = Sale.last.purchase_count*Sale.last.item_price
    else
      gross_income_upload = 0
    end
  end

  def all_time_gross
    if Sale.any?
      all_time_gross = 0
      sales = Sale.all
      sales.each do |sale|
        all_time_gross = all_time_gross + sale.purchase_count*sale.item_price
      end
    else
      all_time_gross = 0
    end
  end

end

用户信息视图（_user_info.html.erb）：

<h1>Gross income uploaded: R$<%= gross_income_upload %></h1>
<h2>All time gross: R$<%= all_time_gross %></h2>

变量“gross_income_upload”中的最后总收入是最后一次销售的收入，在函数中计算如下：

gross_income_upload = Sale.last.purchase_count*Sale.last.item_price

上传的总收入还可以，“gross_income_upload”为“R$20.0”，在视图中呈现良好。

Gross income uploaded: R$20.0

另一个值“all_time_gross”不合适。

预期的结果是循环内的计算，'all_time_gross' 是由 sale purchase_count 乘以数据库中每个条目的 item_price 的总和：

sales.each do |sale|
    all_time_gross = all_time_gross + sale.purchase_count*sale.item_price

视图中的值不是整数，而是 Active Record 数组数据：

All time gross: R$[#<Sale id: 1, purchaser_name: "João Silva", item_description: "R$10 off R$20 of food", item_price: 10.0, purchase_count: 2, merchant_address: "987 Fake St", merchant_name: "Bob's Pizza\n", created_at: "2021-08-30 03:11:57.639361000 +0000", updated_at: "2021-08-30 03:11:57.639361000 +0000">, #<Sale id: 2, purchaser_name: "Amy Pond", item_description: "R$30 of awesome for R$10", item_price: 10.0, purchase_count: 5, merchant_address: "456 Unreal Rd", merchant_name: "Tom's Awesome Shop\n", created_at: "2021-08-30 03:11:58.496490000 +0000", updated_at: "2021-08-30 03:11:58.496490000 +0000">, #<Sale id: 3, purchaser_name: "Marty McFly", item_description: "R$20 Sneakers for R$5", item_price: 5.0, purchase_count: 1, merchant_address: "123 Fake St", merchant_name: "Sneaker Store Emporium\n", created_at: "2021-08-30 03:11:58.593304000 +0000", updated_at: "2021-08-30 03:11:58.593304000 +0000">, #<Sale id: 4, purchaser_name: "Snake Plissken", item_description: "R$20 Sneakers for R$5", item_price: 5.0, purchase_count: 4, merchant_address: "123 Fake St", merchant_name: "Sneaker Store Emporium\n", created_at: "2021-08-30 03:11:58.696714000 +0000", updated_at: "2021-08-30 03:11:58.696714000 +0000">]

Answer 1

Ruby 隐式返回方法中最后一个操作的结果。在all_time_gross 中，当有Sale 对象时，您正在设置一个变量，但实际上并没有返回它。该执行路径中的最后一个操作是sales.each，它（当给定一个块时）在迭代后返回枚举的集合/范围：

def all_time_gross
  if Sale.any?
    ...
    # The result of this is *not* all_time_gross
    sales.each do |sale|
      all_time_gross = all_time_gross + sale.purchase_count*sale.item_price
    end
  else
    ...
  end
end

这就是为什么您最终会得到一组 Sale 对象。在非常简单的层面上，您的问题的解决方案是确保 all_time_gross 是您方法中的最后一件事：

def all_time_gross
  if Sale.any?
    all_time_gross = 0
    sales = Sale.all
    sales.each do |sale|
      all_time_gross = all_time_gross + sale.purchase_count*sale.item_price
    end
  else
    all_time_gross = 0
  end
  all_time_gross # ensures value returned
end

但是，您的方法（实际上两者）可以大大简化，从而减少此类错误蔓延的可能性。对于all_time_gross，您可以使用.sum 方法来处理计算：

def all_time_gross
  # Just bail out if there's nothing to do
  return 0 unless Sale.any?

  # Prefer .find_each over .all.each because it batches queries
  Sale.find_each.sum { |sale| sale.purchase_count * sale.item_price } 
end

在gross_income_upload 中，您也不需要为变量赋值，因为您只是立即返回结果。相反，您可以执行以下操作：

def gross_income_upload
  return 0 unless Sale.any?

  # In your version, calling Sale.last twice causes two queries
  # for the same object
  sale = Sale.last
  sale.purchase_count * sale.item_price
end