Rails 3 - 列出當前用戶還不是成員的組

Question

我正在將應用程序從php轉換爲rails，而且我仍然在學習圍繞rails和AR的方式。我想列出當前用戶還不是成員的組。

失敗的方法：

1. Cause.select('causes.*').joins(:users).group('causes.id').where("cause_user_memberships.user_id NOT IN (?)", current_user.id)

2. Cause.select('causes.*').joins(:users).group('causes.id').where("cause_user_memberships.user_id NOT IN (SELECT cause_user_memberships.cause_id FROM cause_user_memberships WHERE cause_user_memberships.user_id =(?))", current_user.id)

3. 還有更多...

感謝您的幫助！

約車型的一些信息

User.rb（片段）

has_many :cause_user_memberships 
has_many :causes, :through => :cause_user_memberships 

Cause.rb

attr_accessible :title, :location, :description,... 
has_many :cause_user_memberships 
has_many :users, :through => :cause_user_memberships 

Cause_User_Membership.rb（< --probably不我最好的模特名字）

# == Schema Information 
# 
# Table name: cause_user_memberships 
# 
# id   :integer   not null, primary key 
# user_id :integer   not null 
# cause_id :integer   not null 
# created_at :datetime   not null 
# updated_at :datetime   not null 
# 

class CauseUserMembership < ActiveRecord::Base 
    attr_accessible :cause_id, :user_id 

    belongs_to :user 
    belongs_to :cause, :counter_cache => :users_count 
     accepts_nested_attributes_for :cause 
     validates_uniqueness_of :user_id, :scope =>[:cause_id] 

end 

更新：跟進

DERP，你是對的，它的工作！謝謝！

小跟進，查詢時間似乎很長。這是否表明存在問題？我在每張表中的記錄少於20條。 （下面是2個查詢結果，一個包含地理編碼寶石，我打算用，另一個不很抱歉，如果這是一個有些凌亂。）

在Rails控制檯：

Cause Load (1003.0ms) SELECT "causes".* FROM "causes" LEFT JOIN cause_user_memberships ON cause_user_memberships.cause_id = causes.id AND cause_user_memberships.user_id = 1 WHERE (cause_user_memberships.id IS NULL) 

EXPLAIN (34.3ms) EXPLAIN SELECT "causes".* FROM "causes" LEFT JOIN cause_user_memberships ON cause_user_memberships.cause_id = causes.id AND cause_user_memberships.user_id = 1 WHERE (cause_user_memberships.id IS NULL) 

查詢計劃

Hash Right Join (cost=10.45..37.99 rows=1 width=3168) 
    Hash Cond: (cause_user_memberships.cause_id = causes.id) 
    Filter: (cause_user_memberships.id IS NULL) 
    -> Seq Scan on cause_user_memberships (cost=0.00..27.50 rows=7 width=8) 
     Filter: (user_id = 1) 
    -> Hash (cost=10.20..10.20 rows=20 width=3168) 
     -> Seq Scan on causes (cost=0.00..10.20 rows=20 width=3168) 
(7 rows) 

隨着地理編碼本地主機上：

User Load (18.0ms) SELECT "users".* FROM "users" WHERE "users"."id" = 1 LIMIT 1 
    Cause Load (49.6ms) SELECT causes.*, 3958.755864232 * 2 * ASIN(SQRT(POWER(SIN((40.714269 - causes.lat) * PI()/180/2), 2) + COS(40.714269 * PI()/180) * COS(causes.lat * PI()/180) * POWER(SIN((-74.005972 - causes.lng) * PI()/180/2), 2))) AS distance, CAST(DEGREES(ATAN2(RADIANS(causes.lng - -74.005972), RADIANS(causes.lat - 40.714269))) + 360 AS decimal) % 360 AS bearing FROM "causes" LEFT JOIN cause_user_memberships 
ON cause_user_memberships.cause_id = causes.id 
AND cause_user_memberships.user_id = 1 WHERE (causes.lat BETWEEN 36.37231550667456 AND 45.05622249332544 AND causes.lng BETWEEN -79.73435509229111 AND -68.27758890770889 AND 3958.755864232 * 2 * ASIN(SQRT(POWER(SIN((40.714269 - causes.lat) * PI()/180/2), 2) + COS(40.714269 * PI()/180) * COS(causes.lat * PI()/180) * POWER(SIN((-74.005972 - causes.lng) * PI()/180/2), 2))) <= 300) AND (cause_user_memberships.id IS NULL) ORDER BY distance ASC 

Completed 200 OK in 1068ms (Views: 49.1ms | ActiveRecord: 791.4ms)

Answer 1

在軌joins(:symbol)語句變成一個內連接因而不適合用於尋找不存在的關係。您可以手工編寫，以使其左聯接聯接：

Cause.joins(""" 
    LEFT JOIN cause_user_memberships 
    ON cause_user_memberships.cause_id = causes.id 
    AND cause_user_memberships.user_id = #{current_user.id} 
""").where("cause_user_memberships.id IS NULL") 

修訂

內加入將防止一組原因的創造，不具有映射給定的用戶。例如：

Causes 
id | name 
============ 
1 | Cause 1 
2 | Cause 2 

Users 
id | name 
============ 
1 | User 1 

Cause User Memberships 
id | user_id | cause_id 
======================= 
1 | 1  | 1 

SELECT * 
FROM causes 
INNER JOIN cause_user_memberships 
ON cause_user_memberships.cause_id = causes.id 
AND cause_user_memberships.user_id = 1 

將返回

causes.id | causes.name | cause_user_memberships.id | cause_user_memberships.user_id | cause_user_memberships.cause_id 
=========================================================================================================================== 
1   | Cause 1  | 1       | 1        | 1 

，你不能爲了弄清用戶不屬於原因執行任何更多的邏輯。

SELECT * 
FROM causes 
LEFT JOIN cause_user_memberships 
ON cause_user_memberships.cause_id = cause.id 
AND cause_user_memberships.user_id = 1 

將返回

causes.id | causes.name | cause_user_memberships.id | cause_user_memberships.user_id | cause_user_memberships.cause_id 
=========================================================================================================================== 
1   | Cause 1  | 1       | 1        | 1 
2   | Cause 2  | null      | null       | null 

在這種情況下，所有的原因將獲得一排它們是否匹配在cause_user_memberships表或不記錄。現在，您可以申請附加條件梳理出用戶不屬於（其中cause_user_membership.id爲空）的原因。