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Ruby on rails 通过Rails 3中的关联_Ruby On Rails_Ruby On Rails 3.2 - Fatal编程技术网

Ruby on rails 通过Rails 3中的关联

Ruby on rails 通过Rails 3中的关联,ruby-on-rails,ruby-on-rails-3.2,Ruby On Rails,Ruby On Rails 3.2,我在rails 3中遇到了一些问题。我得到了一个“nil:NilClass”错误的未定义方法“full_name” 查看 <% @comparison.each do |v| %> <%= link_to v.trial.user.full_name, trial_trials_path %> <% end %> 控制器 class SearchController < ApplicationController def search

我在rails 3中遇到了一些问题。我得到了一个“nil:NilClass”错误的未定义方法“full_name”

查看

<% @comparison.each do |v| %>
<%= link_to v.trial.user.full_name, trial_trials_path %>
<% end %>

控制器

class SearchController < ApplicationController

  def search
   @comparison = Result.where('variety_id' => [params[:variety_one], params[:variety_two]], 'year' => params[:year]).
   joins(:trial).where('trials.irrigated' => params[:irrigated], 'trials.region_id' => params[:regions])
  end
class User < ActiveRecord::Base

  has_many :trials

  def full_name
    "#{first_name} #{last_name}"
  end

end
class SearchController[params[:variation\u one],params[:variation\u two],'year'=>params[:year])。
联接(:trial).where('trials.irlivated'=>params[:irlivated],'trials.region_id'=>params[:regions])
结束
结果

class Result < ActiveRecord::Base

  belongs_to :trial, :primary_key => 'trial_id'
  has_many :users, :through => :trial

end
类结果'trial\u id'
拥有多个:用户,:至=>:试用版
结束
用户

class SearchController < ApplicationController

  def search
   @comparison = Result.where('variety_id' => [params[:variety_one], params[:variety_two]], 'year' => params[:year]).
   joins(:trial).where('trials.irrigated' => params[:irrigated], 'trials.region_id' => params[:regions])
  end
class User < ActiveRecord::Base

  has_many :trials

  def full_name
    "#{first_name} #{last_name}"
  end

end
class用户
试用版

class Trial < ActiveRecord::Base

  has_many :results
  belongs_to :user, :primary_key => 'user_id'

end
classtrial'user\u id'
结束

当您调用
v.trial.user.full\u name
时-如果Trail没有用户,它将尝试调用nil对象的全名,这将导致您看到的错误


尝试进行一些调试,找出对象为零的原因

为基兰干杯。你是对的,有些条目是空的,但是当我执行“除非v.trial.user.full_name.blank”时,我会得到相同的错误。你需要执行除非v.trial.userbrilliant。再次感谢。