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我通过修改参数而不是在表单中创建 hidden_​​field 来设置用户。据我了解，这是一种更安全的批量分配方式。

  def update
    @exercise = Exercise.find(params[:id])

    #this is the important part
    if params[:exercise][:log_entries_attributes].present?
      params[:exercise][:log_entries_attributes].each do |value|
        value[1].merge!(:user_id => current_user.id)
      end
    end
    #end the important part

    respond_to do |format|
      if @exercise.update_attributes(params[:exercise])
        format.html { redirect_to_back_or_default @exercise, notice: "Exercise was successfully updated." }
        format.mobile { redirect_to @exercise, notice: 'Exercise was successfully updated.' }
        format.json { head :ok }
      else
        format.html { render action: "edit" }
        format.mobile { redirect_to @exercise, notice: "#{@exercise.errors.full_messages.to_sentence}" }
        format.json { render json: @exercise.errors, status: :unprocessable_entity }
      end
    end
  end

在我的规范中，我有以下内容：

describe "with log_entry_attributes" do
  it "updates log_entries_attributes and sets user" do
    exercise = FactoryGirl.create(:exercise)
    log_entry = FactoryGirl.build(:log_entry)
    exercise.log_entries << log_entry
    exercise.save
    controller.stub(:current_user).and_return(@user)
    put :update, :id => exercise.id, :exercise => FactoryGirl.build(:exercise, "log_entries_attributes" => {":0" => {"reps" => "5", "weight" => "5"}}).attributes.symbolize_keys
    assigns(:exercise).log_entries.first.user.should eq(@user)
  end
end

我明白了undefined method user for nil:NilClass。我想我知道为什么我会得到未定义的方法用户。只是没有办法通过分配获得关联。我不确定如何测试 user_id 是否通过 current_user 正确设置。有什么帮助吗？

4

1 回答 1

1

使用模拟对象：

exercise = double "exercise"
Exercise.should_receive(:find).and_return(exercise)

并测试：

exercise.should_receive(:update_attributes).with(correct_params)
于 2012-08-22T09:28:50.300 回答