模型法或控制器实例变量，Ruby on Rails的3

Question

如果我有象模型的方法：

def favoured_users 
self.followers.limit(5).order("created_at") 
end 

与像一个视图块：

<% 5.times do |i| %> 
    <li><%= @user.favoured_users[i].name %></li> 
<% end %> 

...将我被调用favoured_user方法五次，每次请求5个用户，最终得到25个被调用的用户？

我只是想知道我是否应该把favoured_users的结果在一个变量在我的控制器来代替：

@favoured_users = @user.followers.limit(5).order("created_at") 

那是要到服务器更少电话？

**编辑**

我不知道这是否意味着该值从缓存中来，看来它是（缓存BCOS，但我不知道那是什么意思），但我还没有明确告诉它，我必须做什么，以确保它不来回回缓存：

User Load (0.6ms) SELECT `users`.* FROM `users` INNER JOIN `relationships` ON `users`.`id` = `relationships`.`followed_id` WHERE `relationships`.`follower_id` = 1 ORDER BY full_name, created_at LIMIT 5 
CACHE (0.0ms) SELECT `users`.* FROM `users` INNER JOIN `relationships` ON `users`.`id` = `relationships`.`followed_id` WHERE `relationships`.`follower_id` = 1 ORDER BY full_name, created_at LIMIT 5 
CACHE (0.0ms) SELECT `users`.* FROM `users` INNER JOIN `relationships` ON `users`.`id` = `relationships`.`followed_id` WHERE `relationships`.`follower_id` = 1 ORDER BY full_name, created_at LIMIT 5 
CACHE (0.0ms) SELECT `users`.* FROM `users` INNER JOIN `relationships` ON `users`.`id` = `relationships`.`followed_id` WHERE `relationships`.`follower_id` = 1 ORDER BY full_name, created_at LIMIT 5 
CACHE (0.0ms) SELECT COUNT(*) FROM `users` INNER JOIN `relationships` ON `users`.`id` = `relationships`.`follower_id` WHERE `relationships`.`followed_id` = 1 
User Load (0.5ms) SELECT `users`.* FROM `users` INNER JOIN `relationships` ON `users`.`id` = `relationships`.`follower_id` WHERE `relationships`.`followed_id` = 1 ORDER BY full_name, created_at LIMIT 5 
CACHE (0.0ms) SELECT `users`.* FROM `users` INNER JOIN `relationships` ON `users`.`id` = `relationships`.`follower_id` WHERE `relationships`.`followed_id` = 1 ORDER BY full_name, created_at LIMIT 5 
CACHE (0.0ms) SELECT `users`.* FROM `users` INNER JOIN `relationships` ON `users`.`id` = `relationships`.`follower_id` WHERE `relationships`.`followed_id` = 1 ORDER BY full_name, created_at LIMIT 5 
CACHE (0.0ms) SELECT `users`.* FROM `users` INNER JOIN `relationships` ON `users`.`id` = `relationships`.`follower_id` WHERE `relationships`.`followed_id` = 1 ORDER BY full_name, created_at LIMIT 5 

是值从缓存中返回？

编辑我不知道如何从我的观点访问变量，我有方法按塞比的编辑和在我看来，我想：

<% @user.favoured_followers do %> 

    <li><%= @favoured.first.username unless @favoured.first.blank? %></li> 
    <li><%= @favoured.second.username unless @favoured.second.blank? %></li> 
    <li><%= @favoured.third.username unless @favoured.third.blank? %></li> 
    <li><%= @favoured.fourth.username unless @favoured.fourth.blank? %></li> 
    <li><%= @favoured.fifth.username unless @favoured.fifth.blank? %></li> 

<% end %> 

没有被退回？

Answer 1

如果你看看你的日志，你将能够验证它是否打到缓存或不是每次调用函数。这似乎没有达到缓存，因此加载关注者的查询将为每个用户重复5次。即对于每个@user，您将触发该查询5次。因此，将值缓存在变量中肯定是一个更好的主意。

编辑： 你可以改变你的模型方法是这样的：

def favoured_users 
    @favoured ||= self.followers.limit(5).order("created_at") 
end 

的@favoured变量将要创建的第一次这就是所谓的，然后任何后续调用将不退还查询被解雇。

您的视图代码应保持不变。即：

<% 5.times do |i| %> 
    <li><%= @user.favoured_users[i].name %></li> 
<% end %>  

Answer 2

如果我必须做的..我会做下面的方式

内部控制

@favoured_users = @user.followers.limit(5).order(:created_at) 

内景

<% @favoured_users do |user| %> 
    <li><%= user.name %></li> 
<% end %> 
<% (5 - @favoured_users.count).times do%> 
<li>&nbsp;</li> 
<% end %>