在rails中跳过JSON格式生成脚手架

Question

当您使用命令生成rails脚手架时,rails g scaffold Thing有任何方法可以避免令人讨厌

respond_to do |format|
  format.html # index.html.erb
  format.json { render json: @things }
end

你控制器里的东西？

我正在尝试在Rails上教一个类,我想首先让它们生成一个脚手架,但是所有json格式化它都比它需要的要复杂得多.如果他们能够生成一个创建这样的控制器的脚手架,我会更高兴:

class ThingsController < ApplicationController

  def index
    @things = Thing.all
  end

  def show
    @thing = Thing.find(params[:id])
  end

  def new
    @thing = Thing.new
  end

  def edit
    @thing = Thing.find(params[:id])
  end

  def create
    @thing = Thing.new(params[:thing])
      if @thing.save
        redirect_to @thing, notice: 'Thing was successfully created.'
      else
        render: "new" 
      end
    end
  end

  def update
    @thing = Thing.find(params[:id])
      if @thing.update_attributes(params[:thing])
        redirect_to @thing, notice: 'Thing was successfully updated.'
      else
        render: "edit" 
      end
    end
  end

  def destroy
    @thing = Thing.find(params[:id])
    @thing.destroy
    redirect_to things_url
  end
end

Answer 1

注释掉jbuilder你的Gemfile和你的respond_to块中的gem 将不会被生成.

Answer 2

只需克隆文件

https://github.com/rails/rails/blob/v5.2.2/railties/lib/rails/generators/rails/scaffold_controller/scaffold_controller_generator.rb

到你的

lib/rails/generators/rails/scaffold_controller/templates/controller.rb

应用程序中的路径并自定义您想要的内容.此外,您可以编写自己的脚手架生成器(http://guides.rubyonrails.org/generators.html).