我正在开发一个允许会员进行调查的应用程序(会员与Response有一对多的关系).Response保存member_id,question_id及其答案.
调查全部或全部提交,因此如果该成员的响应表中有任何记录,则他们已完成调查.
我的问题是,如何重新编写下面的查询,以便它实际工作?在SQL中,这将是EXISTS关键字的主要候选者.
def surveys_completed
members.where(responses: !nil ).count
end
Run Code Online (Sandbox Code Playgroud) 这是我的控制器
@post = Post.joins(:customers).select("customers.*,posts.*").find params[:id]
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我的帖子模型
belongs_to :customer
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我的客户模特
has_many :posts
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我得到的错误是
Association named 'customers' was not found on Post; perhaps you misspelled it?
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这是我的控制器输出:
Processing by PostsController#show as */*
Parameters: {"id"=>"6"}
Post Load (0.5ms) SELECT "posts".* FROM "posts" WHERE "posts"."id" = $1 LIMIT 1 [["id", "6"]]
Completed 500 Internal Server Error in 113ms
ActiveRecord::ConfigurationError (Association named 'customers' was not found on Post; perhaps you misspelled it?):
app/controllers/posts_controller.rb:16:in `show'
Run Code Online (Sandbox Code Playgroud) 我找到了一个答案,其中有一些可用的having例子可以找到有n孩子的父母,但同样不能用于找到没有孩子的父母(大概是因为连接不包括他们).
scope :with_children, joins(:children).group("child_join_table.parent_id").having("count(child_join_table.parent_id) > 0")
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谁能指出我正确的方向?
如果我有一个模型Person,has_many Vehicles每个模型都Vehicle可以是类型,car或者motorcycle我如何查询所有拥有汽车的人和所有拥有摩托车的人?
我不认为这些是正确的:
Person.joins(:vehicles).where(vehicle_type: 'auto')
Person.joins(:vehicles).where(vehicle_type: 'motorcycle')
Run Code Online (Sandbox Code Playgroud) 我的模特在这里:
class User < ActiveRecord::Base
has_many :bookmarks
end
class Topic < ActiveRecord::Base
has_many :bookmarks
end
class Bookmark < ActiveRecord::Base
belongs_to :user
belongs_to :topic
attr_accessible :position
validates_uniqueness_of :user_id, :scope => :topic_id
end
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我想topics用current_user相关的for 来获取所有内容bookmark.ATM,我这样做:
Topic.all.each do |t|
bookmark = t.bookmarks.where(user_id: current_user.id).last
puts bookmark.position if bookmark
puts t.name
end
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这很难看并且做了太多的数据库查询.我想要这样的东西:
class Topic < ActiveRecord::Base
has_one :bookmark, :conditions => lambda {|u| "bookmarks.user_id = #{u.id}"}
end
Topic.includes(:bookmark, current_user).all.each do |t| # this must also includes topics without bookmark …Run Code Online (Sandbox Code Playgroud)