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我正在开发一个允许用户对食谱进行评分的项目。我制作了一个 User 和 Recipe 共享的简单连接表 (RecipeRating)，但 ActiveRecord 无法保存。

@rating = @user.recipe_ratings.where(:recipe_id => @recipe.id).first
@rating.rating = params[:rating]
@rating.save

产量：

ActiveRecord::StatementInvalid in UsersController#rate_recipe
PG::Error: ERROR:  zero-length delimited identifier at or near """"
LINE 1: ...013-08-07 14:28:39.965953' WHERE "recipe_ratings"."" IS NULL
                                                             ^
: UPDATE "recipe_ratings" SET "rating" = 3, "updated_at" = '2013-08-07 14:28:39.965953' WHERE "recipe_ratings"."" IS NULL

显然错误是由WHERE "recipe_ratings"."" IS NULL 什么引起的，我该如何解决？作为参考，创建或查找评级（按预期保存在我的数据库中）没有问题，它只是无法更新它们。

以供参考：

class User < ActiveRecord::Base
  ...
  has_many :recipe_ratings
end

class Recipe < ActiveRecord::Base
  ...
  has_many :recipe_ratings
end

class RecipeRating < ActiveRecord::Base
  belongs_to :recipe
  belongs_to :user
  attr_accessible :comment, :rating, :user, :recipe#, :user_id, :recipe_id
end
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1 回答 1

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我通过为 RecipeRating 表提供自己的主键来解决此问题。

于 2013-08-11T05:30:59.890 回答