Ruby-on-rails 生成具有所有 RESTful 功能的控制器

Generate a controller with all the RESTful functions

提问by Barb

I am trying to generate a controller with all the RESTful actions stubbed. I had read at link textthat all I needed to do was to use call the generator with just a controller name I would get just that. So, I ran "script/generate rspec_controller Properties" and I got an empty controller.

我正在尝试生成一个带有所有 RESTful 操作的控制器。我在链接文本中读到，我需要做的就是使用仅使用控制器名称调用生成器，我会得到的。所以，我运行了“脚本/生成 rspec_controller 属性”，我得到了一个空控制器。

Any other suggestions would be greatly appreciated.

任何其他建议将不胜感激。

回答by Marcos Placona

I don't know about an automated way of doing it, but if you do:

我不知道这样做的自动化方式，但如果你这样做：

script/generate controller your_model_name_in_plural new create update edit destroy index show

All of them will be created for you

所有这些都将为您创建

Update for Rails 4

Rails 4 的更新

rails g scaffold_controller Property

回答by gdelfino

In Rails 3 there is also rails generate scaffold_controller .... More info here.

在 Rails 3 中也有rails generate scaffold_controller .... 更多信息在这里。

回答by konung

EDIT(due to some comments): Original question was in 2010 - hence the answer is NOT for RAILS 4 , but for rails 2!!

编辑（由于一些评论）：最初的问题是在 2010 年 - 因此答案不是针对 RAILS 4 ，而是针对 rails 2 ！

try using scaffolding.

尝试使用脚手架。

script/generate scaffold controller Properties

Section of Official docs on Ruby On Rails

Ruby On Rails 官方文档部分

I'm sure you can find more info if you do a google search on rails scaffolding. Hope that helps.

如果您在谷歌搜索 Rails 脚手架，我相信您可以找到更多信息。希望有帮助。

EDIT:For RAILS 4

编辑：对于导轨 4

rails g scaffold_controller Property

rails g scaffold_controller Property

回答by pmargreff

In Rails 4/5 the following command does the trick for me.

在 Rails 4/5 中，以下命令对我有用。

rails g scaffold_controller Property --skip-template-engine

It generated the controller actions but not the view.

它生成了控制器动作而不是视图。

回答by Varus Septimus

Rails 5.1

导轨 5.1

Starting point:

初始点：

You have created a model without a controller, nor views (eg thru: rails generate model category)

您创建了一个没有控制器的模型，也没有视图（例如 thru: rails generate model category）

Objective:

客观的：

Upgrade it to a full RESTful resource

将其升级为完整的 RESTful 资源

Command:

命令：

rails generate scaffold_controller category

rails generate scaffold_controller category

It stubs out a scaffolded controller, its seven RESTful actions and related views. (Note: You can either pass the model name CamelCased or under_scored.)

它生成了一个脚手架控制器、七个 RESTful 操作和相关视图。（注意：您可以传递模型名称 CamelCased 或 under_scored。）

Output:

输出：

varus@septimusSrv16DEV4:~/railsapps/dblirish$ rails generate scaffold_controller category
Running via Spring preloader in process 45681
      create  app/controllers/categories_controller.rb
      invoke  erb
      create    app/views/categories
      create    app/views/categories/index.html.erb
      create    app/views/categories/edit.html.erb
      create    app/views/categories/show.html.erb
      create    app/views/categories/new.html.erb
      create    app/views/categories/_form.html.erb
      invoke  test_unit
      create    test/controllers/categories_controller_test.rb
      invoke  helper
      create    app/helpers/categories_helper.rb
      invoke    test_unit
      invoke  jbuilder
      create    app/views/categories/index.json.jbuilder
      create    app/views/categories/show.json.jbuilder
      create    app/views/categories/_category.json.jbuilder

回答by molf

You're looking for scaffolding.

您正在寻找脚手架。

Try:

尝试：

script/generate scaffold Property

This will give you a controller, a model, a migration and related tests. You can skip the migration with the option --skip-migration. If you don't want the others, you'll have to delete them yourself. Don't worry about overwriting existing files, that won't happen unless you use --force.

这将为您提供控制器、模型、迁移和相关测试。您可以使用选项跳过迁移--skip-migration。如果您不想要其他人，则必须自己删除它们。不要担心覆盖现有文件，除非您使用--force.

As klew points out in the comments, this also defines the method bodiesfor you, not just the names. It is very helpful to use as a starting point for your REST controller.

正如 klew 在评论中指出的那样，这也为您定义了方法主体，而不仅仅是名称。用作 REST 控制器的起点非常有帮助。

回答by Tom Hammond

In Rails 4 it's rails g controller apps new create update edit destroy show index

在 Rails 4 中 rails g controller apps new create update edit destroy show index

Or rails generate controller apps new create update edit destroy show indexif you want to write out the full term :).

或者rails generate controller apps new create update edit destroy show index如果你想写出完整的术语:)。

回答by stephenmurdoch

script/generate rspec_scaffold Property

脚本/生成 rspec_scaffold 属性

回答by rfunduk

There's no way (that I know of? that is documented?) to stub out a controller except through scaffolding. But you could do:

除了通过脚手架之外，没有办法（我知道的？有记录的？）来消除控制器。但你可以这样做：

script/generate controller WhateverController new create edit update destroy show

回答by philippinedev

One solution is to create a script that accepts one parameter, the controller name, and let the script type the whole command for you.

一种解决方案是创建一个脚本，该脚本接受一个参数，即控制器名称，并让脚本为您键入整个命令。

---

---

1. Create a new file, say, railsgcontroller
2. Make it executable and save it on path
3. Run it like: $ railsgcontroller Articles

1. 创建一个新文件，比如 railsgcontroller
2. 使其可执行并将其保存在路径上
3. 像这样运行： $ railsgcontroller 文章

---

---

die () {
    echo "Please supply new rails controller name to generate."
    echo >&2 "$@"
    exit 1
}

[ "$#" -eq 1 ] || die "1 argument required, $# provided"

rails g controller "" new create update edit destroy show index