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我正在将应用程序从 php 转换为 rails，并且我仍在学习有关 rails 和 AR 的方法。

简单地说：我想列出当前用户还不是成员的组。

失败的方法：

  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. 还有很多...

谢谢你的帮助！

关于模型的一些信息

用户.rb（片段）

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

原因.rb

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

Cause_User_Membership.rb（<--可能不是我最好的模型名称）

    # == 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：跟进

Derp，你说得对，它奏效了！谢谢！

小跟进，查询时间似乎很长。这是否表明有问题？我在每个表中只有不到 20 条记录。（以下是 2 个查询结果，一个包含我打算使用的地理编码器 gem，另一个不包含。如果有点乱，请见谅。）

在 Rails 控制台中： 

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)

查询计划

 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)

在本地主机上使用地理编码器： 

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)

4

1 回答 1

2

在 Rails 中，该joins(:symbol)语句变成了不适合查找不存在的关系的内部连接。您可以手动编写连接以使其成为左连接：

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")

更新

内部连接将阻止创建一组没有到给定用户的映射的原因。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

将返回

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

为了找出用户不属于的原因，您无法再执行任何逻辑。

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

将返回

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

在这种情况下，无论是否匹配表中的记录，所有原因都将获得一行cause_user_memberships。现在您可以应用附加条件来梳理出用户不属于的原因（其中 cause_user_membership.id 为空）。

于 2012-12-04T20:41:43.730 回答