我期望从同一视图中显示的活动记录数据计算的助手中的两个值，'gross_income_upload'和'all_time_gross'。总收入是可以的，但是没有显示总收入整数，而是显示活动记录数组数据。
有一个helper函数和一个.erb视图，代码如下:
应用程序帮助器(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，在函数中按以下方式计算:
gross_income_upload = Sale.last.purchase_count*Sale.last.item_price
上传的总收入是OK的，'gross_income_upload'是'R$20.0'，它在视图中被很好地呈现。
Gross income uploaded: R$20.0
不合适的是另一个值'all_time_gross'。
期望的结果是循环内的计算，'all_time_gross'是由database中每个条目的sale purchase_count乘以item_price的总和来测量的:
sales.each do |sale|
all_time_gross = all_time_gross + sale.purchase_count*sale.item_price
视图中的值是Active Record数组数据，而不是整数:
All time gross: R$[#<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 Pizzan", 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 Shopn", 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 Emporiumn", 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 Emporiumn", created_at: "2021-08-30 03:11:58.696714000 +0000", updated_at: "2021-08-30 03:11:58.696714000 +0000">]
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，您设置了一个变量，但实际上并不返回它。该执行路径中的最后一个操作是sales.each，它(当给定一个块时)在迭代后返回枚举集/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
这就是为什么你结束了你的Sale对象的数组。在一个非常简单的层面上，你的问题的解决方案是确保all_time_gross是你的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
然而，您的方法(实际上两者都是)可以大大简化，从而减少了此类错误潜入的可能性。在all_time_gross的情况下，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，你可以这样做:
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