模型法或控制器實例變量，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 

與像一個視圖塊：

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

...將我被調用favoured_user方法五次，每次請求5個用戶，最終得到25個被調用的用戶？

我只是想知道我是否應該把favoured_users的結果在一個變量在我的控制器來代替：

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

那是要到服務器更少電話？

**編輯**

我不知道這是否意味着該值從緩存中來，看來它是（緩存BCOS，但我不知道那是什麼意思），但我還沒有明確告訴它，我必須做什麼，以確保它不來回回緩存：

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 

是值從緩存中返回？

編輯我不知道如何從我的觀點訪問變量，我有方法按塞比的編輯和在我看來，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 %> 

沒有被退回？

Answer 1

如果你看看你的日誌，你將能夠驗證它是否打到緩存或不是每次調用函數。這似乎沒有達到緩存，因此加載關注者的查詢將爲每個用戶重複5次。即對於每個@user，您將觸發該查詢5次。因此，將值緩存在變量中肯定是一個更好的主意。

編輯： 你可以改變你的模型方法是這樣的：

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

的@favoured變量將要創建的第一次這就是所謂的，然後任何後續調用將不退還查詢被解僱。

您的視圖代碼應保持不變。即：

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

Answer 2

如果我必須做的..我會做下面的方式

內部控制

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

內景

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