Question

我有这个网站，用户可以有很多宠物。这些宠物列在那里的配置文件中，点击宠物将它们带到宠物配置文件中，在那里他们可以选择编辑宠物。加载宠物的表格效果很好，它为每个人提供了正确的信息，但随后出现了更新问题。单击编辑表单上的提交会导致所有更新转到他们拥有的第一个宠物。比如，如果宠物的ID为2,5和11，而5或11的更新则转为2.我不太确定发生了什么。

以下是PetsController的编辑和更新操作：

#GET to /users/:user_id/pet/edit
def edit
  @user = User.find( params[:user_id] )
  @pet = Pet.find_by_id( params[:id] )
end

#PUT to /users/:user_id/pet/
def update
  #Retrieve user from database
  @user = User.find( params[:user_id] )
  #Retrieve user's pet
  @pet = @user.pets.find_by( params[:id] )
  #Mass assign edited pet attributes and save (update)
  if @pet.update_attributes(pet_params)
    flash[:success] = "Pet updated!"
    #Redirect user to profile page
    redirect_to user_path(id: params[:user_id] )
  else 
    render action: :edit
    Rails.logger.info(@pet.errors.messages.inspect)
  end
end

我尝试过不同的寻找用户宠物的方法，例如

@user.pet(params[:id])

这导致了一个错误，它说找不到没有身份证的宠物，所以接下来我尝试了

@user.pet.find_by_id(params[:id])

但是这导致update_attribute没有被定义，所以我认为我不应该使用find_by_id

我可能还有另一种方式吗？我认为错误是宠物的身份证没有被正确传递。如果是这样，有什么可以解决这个问题？我尝试将ID作为隐藏字段从编辑表单中传递，但这并不起作用。

这是当用户尝试更新宠物时在终端中发生的事情

Processing by PetsController#update as HTML
Parameters: {"utf8"=>"✓",   "authenticity_token"=>"0tC+qRK01CMCEZWbDtJyZ4QNmXQxnaAPKT+anRKGKh5hTWekTbEHzAd+8 +YOVIWBB0/imVrMVNcpkeBR18SHNw==", "pet"=>{"color"=>"Darigan",  "species"=>"Yurble", "gender"=>"Male", "level"=>"25", "hp"=>"50",  "strength"=>"50", "defence"=>"50", "movement"=>"50", "uc"=>"1", "rw"=>"0", "rn"=>"0", "uft"=>"1", "ufa"=>"0", "description"=>"uiy7i77i7i", "id"=>"5",  "name"=>"Fouany", "hsd"=>"150"}, "user_id"=>"5"}
User Load (0.3ms)  SELECT  "users".* FROM "users" WHERE "users"."id" = ?  LIMIT ?  [["id", 5], ["LIMIT", 1]]
Pet Load (0.3ms)  SELECT  "pets".* FROM "pets" WHERE "pets"."user_id" = ? LIMIT ?  [["user_id", 5], ["LIMIT", 1]]
(0.1ms)  begin transaction
Pet Exists (0.2ms)  SELECT  1 AS one FROM "pets" WHERE "pets"."name" = ? AND ("pets"."id" != ?) LIMIT ?  [["name", "Fouany"], ["id", 2], ["LIMIT", 1]]
(0.1ms)  rollback transaction
Rendering pets/edit.html.erb within layouts/application
User Load (0.3ms)  SELECT  "users".* FROM "users" WHERE "users"."id" = ? ORDER BY "users"."id" ASC LIMIT ?  [["id", 5], ["LIMIT", 1]]
Rendered pets/edit.html.erb within layouts/application (12.4ms)
Profile Load (0.4ms)  SELECT  "profiles".* FROM "profiles" WHERE "profiles"."user_id" = ? LIMIT ?  [["user_id", 5], ["LIMIT", 1]]
{:name=>["has already been taken"], :color=>[], :species=>[], :gender=>[], :level=>[], :hp=>[], :strength=>[], :defence=>[], :movement=>[], :uc=>[], :rw=>[], :rn=>[], :uft=>[], :ufa=>[], :description=>[]}
Completed 200 OK in 202ms (Views: 136.1ms | ActiveRecord: 5.0ms)

正如你所看到的，宠物的身份证从5开始从5回到2，所以也许可以从某个地方开始解决这个问题？

非常感谢任何建议和意见！

Answer 1

@user.pets.find_by(params[:id])

应该是

@user.pets.find_by(id: params[:id])

Answer 2

如果您的用户应该只有一只宠物（路线建议这样），则不需要宠物ID来访问它。

你可以得到宠物

@user = User.find( params[:user_id] )
@pet = @user.pet

在您的代码中查看此评论：

#PUT to /users/:user_id/pet/

提到了user_id，但未提到id

Ruby on Rails更新保存到错误的ID

2 个答案: