I wish to create my custom generator, some like run the command "rails generator myScaffold entity field1:String field2:String
and the generator behavior like a normal scaffold, except it will not create the stylesshets, views, and the class of the controller would be customized.
Is it posible with rails? And is it a good and correct thing to do?
thanks!!!
All you need is available here http://guides.rubyonrails.org/generators.html
I would like to create a model called models/thing.rb, a controller called controllers/things_controller.rb, and views called views/things/index.html.erb and views/things/hello.html.erb.
How do I do this from the rails console? I know I have to use rails generate.
Firstly generate the controller with two actions index and hello.
rails g controller things index hello
This command will generate the controller and the views with views/things/index.html.erb and views/things/hello.html.erb.
Then generate the model:
rails g model Thing attribute:type [...]
If you want to know more about rails generate command, you can take a look at A Guide to The Rails Command Line
My setup: Rails 3.0.9, Ruby 1.9.2
I wish to generate a scaffold only for the create action, what's the syntax for that? I'm guessing it's something I append to
rails g scaffold Project name:string ...
I don't believe you can. I'm looking at the generator and I see nothing related to using an option to limit the template.
Generator
Template
Rails scaffolding is a very basic tool, it's not meant to be relied on, once you get a handle on the way rails works. However, you can use Ryan Bates' nifty-generator gem to generate scaffolds with greater control. Example:
rails g nifty:scaffold post name:string index new edit
Learn more here. https://github.com/ryanb/nifty-generators
I have rails 2.3 and ruby 1.8.7
I want to create only a controller and view files for it. Is there any command in rails2.3 that will help me in doing this. I know this can be done in rails 3, but is there a way to do this in rails 2.3
Just say script/generate controller [controller_name] [action_name_1] [action_name_2] ... in the Rails directory to create a controller with the given actions and corresponding view files.
I am trying to generate a controller with all the RESTful actions stubbed. I had read at Wikibooks - Ruby on Rails that all I needed to do was to call the generator with the controller name and I would get just that. So, I ran script/generate rspec_controller Properties but got an empty controller.
Any other suggestions would be greatly appreciated.
I don't know about an automated way of doing it, but if you do:
script/generate controller your_model_name_in_plural new create update edit destroy index show
All of them will be created for you
Update for Rails 4
rails g scaffold_controller Property
In Rails 3 there is also rails generate scaffold_controller .... More info here.
EDIT(due to some comments) : Original question was in 2010 - hence the answer is NOT for RAILS 4 , but for rails 2!!
try using scaffolding.
script/generate scaffold controller Properties
Section of Official docs on Ruby On Rails
I'm sure you can find more info if you do a google search on rails scaffolding. Hope that helps.
EDIT:
For RAILS 4
rails g scaffold_controller Property
In Rails 4/5 the following command does the trick for me.
rails g scaffold_controller Property --skip-template-engine
It generated the controller actions but not the view.
Rails 5.1
Starting point:
You have created a model without a controller, nor views (eg thru: rails generate model category)
Objective:
Upgrade it to a full RESTful resource
Command:
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.)
Output:
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
You're looking for scaffolding.
Try:
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.
As klew points out in the comments, this also defines the method bodies for you, not just the names. It is very helpful to use as a starting point for your REST controller.
In Rails 4 it's rails g controller apps new create update edit destroy show index
Or rails generate controller apps new create update edit destroy show index if you want to write out the full term :).
script/generate rspec_scaffold Property
There's no way (that I know of? that is documented?) to stub out a controller except through scaffolding. But you could do:
script/generate controller WhateverController new create edit update destroy show
One solution is to create a script that accepts one parameter, the controller name, and let the script type the whole command for you.
Create a new file, say, railsgcontroller
Make it executable and save it on path
Run it like:
$ railsgcontroller Articles
die () {
echo "Please supply new rails controller name to generate."
echo >&2 "$#"
exit 1
}
[ "$#" -eq 1 ] || die "1 argument required, $# provided"
rails g controller "$1" new create update edit destroy show index