的has_many通過與秩序不工作

Question

我試圖通過EventViews到列表中的用戶，並責令其選擇，但我不斷收到一個錯誤：

Unknown column 'views_count' in 'order clause': SELECT `users`.`id` AS t0_r0, `users`.`email` AS t0_r1, `users`.`encrypted_password` AS t0_r2, `users`.`reset_password_token` AS t0_r3, `users`.`reset_password_sent_at` AS t0_r4, `users`.`remember_created_at` AS t0_r5, `users`.`sign_in_count` AS t0_r6, `users`.`current_sign_in_at` AS t0_r7, `users`.`last_sign_in_at` AS t0_r8, `users`.`current_sign_in_ip` AS t0_r9, `users`.`last_sign_in_ip` AS t0_r10, `users`.`created_at` AS t0_r11, `users`.`updated_at` AS t0_r12, `users`.`user_type` AS t0_r13, `event_views`.`id` AS t1_r0, `event_views`.`event_id` AS t1_r1, `event_views`.`ip` AS t1_r2, `event_views`.`created_at` AS t1_r3, `event_views`.`updated_at` AS t1_r4, `event_views`.`session_id` AS t1_r5, `event_views`.`is_unique` AS t1_r6, `event_views`.`user_agent` AS t1_r7, `event_views`.`promoter_id` AS t1_r8 FROM `users` LEFT OUTER JOIN `events` ON `events`.`user_id` = `users`.`id` LEFT OUTER JOIN `event_views` ON `event_views`.`event_id` = `events`.`id` ORDER BY views_count desc 

這裏是我的代碼：

User.all.includes(:event_views).select('user.*, count(event_views.id) as "views_count"').order("views_count desc").each 

#models 

User 
has_many :event_views, :through => :events 

Events 
has_many :event_views 

EventView 
#nothing relevant 

他們是所有相關的正常時尚（event_id，user_id等）

Answer 1

這裏有幾點需要記住：

1. 您正在致電includes上User.all這是一個Array，所以這將失敗。
2. 正如@peteralfvin已經指出，你應該有users.*而不是user.*。但是這會是一個問題，因爲你正在使用count聚合函數。這意味着您需要在group by子句中包含users表中的每一列。
3. 在這種情況下，您不應該使用includes，因爲includes將包含查詢中所有包含表中的所有列。這是因爲您正在從users和event_views中選擇列。如果您刪除count(event_views.id) as "views_count"，則會看到只有users表中的指定列將被選中。

話雖這麼說，這個查詢應該爲你工作：

users = User.joins(:event_views).select('users.id, count(event_views.id) as views_count').order("views_count desc") 
# Then access your views_count for user with: 
users.first.views_count 

Answer 2

你並不真的需要做到這一點的SQL查詢，如果你得到的所有記錄無論如何：

users = User.includes(:event_views).all 
users.sort_by!{ |user| user.event_views.count } 

如果你想讓它們以另一種方式排序，只需在最後一行的末尾加上.reverse!即可。