Rails create new MVC - ruby-on-rails

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

Related

How to make database on ruby on rails?

I have just created the project on ruby on rails. I also generated a controller with an instruction on bash, "rails g controller home".
However, I didn't make database. How can I generate database? Please inform me how to make database.
Thank you.
You can generate a database table by typing:
rails g model myTableName
Because your starting out new i think it would be best to try something like:
rails g scaffold Post name:string content:text
This will link the controller, the model(which is the database) and the views together, giving you a better understanding on how the MVC structure works in rails.

Scaffold only a subset of methods in controller

Is there a way to generate a subset of methods in the controller using scaffold instead of the regular scaffold?
For example: Only create new and show?
I am using rails 3, ruby 1.9.2
Thanks
I do not think the default rails scaffold generator supports this in Rails 3.
However, this can be accomplished using the nifty-generators gem:
https://github.com/ryanb/nifty-generators
In plain Rails3, you can do this when you generate the controller:
rails generate controller Artists new create
So, you could just generate the model, and then the controller, and get the overall functionality you were looking for.

Create scaffold by skipping the model

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.

Rails: How to run `rails generate scaffold` when the model already exists?

I'm new to Rails so my current project is in a weird state.
One of the first things I generated was a "Movie" model. I then started defining it in more detail, added a few methods, etc.
I now realize I should have generated it with rails generate scaffold to hook up things like the routing, views, controller, etc.
I tried to generate the scaffolding but I got an error saying a migration file with the same name already exists.
What's the best way for me to create scaffolding for my "Movie" now? (using rails 3)
TL;DR: rails g scaffold_controller <name>
Even though you already have a model, you can still generate the necessary controller and migration files by using the rails generate option. If you run rails generate -h you can see all of the options available to you.
Rails:
controller
generator
helper
integration_test
mailer
migration
model
observer
performance_test
plugin
resource
scaffold
scaffold_controller
session_migration
stylesheets
If you'd like to generate a controller scaffold for your model, see scaffold_controller. Just for clarity, here's the description on that:
Stubs out a scaffolded controller and its views. Pass the model name,
either CamelCased or under_scored, and a list of views as arguments.
The controller name is retrieved as a pluralized version of the model
name.
To create a controller within a module, specify the model name as a
path like 'parent_module/controller_name'.
This generates a controller class in app/controllers and invokes helper,
template engine and test framework generators.
To create your resource, you'd use the resource generator, and to create a migration, you can also see the migration generator (see, there's a pattern to all of this madness). These provide options to create the missing files to build a resource. Alternatively you can just run rails generate scaffold with the --skip option to skip any files which exist :)
I recommend spending some time looking at the options inside of the generators. They're something I don't feel are documented extremely well in books and such, but they're very handy.
Great answer by Lee Jarvis, this is just the command e.g; we already have an existing model called User:
rails g scaffold_controller User
For the ones starting a rails app with existing database there is a cool gem called schema_to_scaffold to generate a scaffold script.
it outputs:
rails g scaffold users fname:string lname:string bdate:date email:string encrypted_password:string
from your schema.rb our your renamed schema.rb. Check it
In Rails 5, you can still run
$rails generate scaffold movie --skip
to create all the missing scaffold files or
rails generate scaffold_controller Movie
to create the controller and view only.
For a better explanation check out rails scaffold
This command should do the trick:
$ rails g scaffold movie --skip
You can make use of scaffold_controller and remember to pass the attributes of the model, or scaffold will be generated without the attributes.
rails g scaffold_controller User name email
# or
rails g scaffold_controller User name:string email:string
This command will generate following files:
create app/controllers/users_controller.rb
invoke haml
create app/views/users
create app/views/users/index.html.haml
create app/views/users/edit.html.haml
create app/views/users/show.html.haml
create app/views/users/new.html.haml
create app/views/users/_form.html.haml
invoke test_unit
create test/controllers/users_controller_test.rb
invoke helper
create app/helpers/users_helper.rb
invoke test_unit
invoke jbuilder
create app/views/users/index.json.jbuilder
create app/views/users/show.json.jbuilder
I had this challenge when working on a Rails 6 API application in Ubuntu 20.04.
I had already existing models, and I needed to generate corresponding controllers for the models and also add their allowed attributes in the controller params.
Here's how I did it:
I used the rails generate scaffold_controller to get it done.
I simply ran the following commands:
rails generate scaffold_controller School name:string logo:json motto:text address:text
rails generate scaffold_controller Program name:string logo:json school:references
This generated the corresponding controllers for the models and also added their allowed attributes in the controller params, including the foreign key attributes.
create app/controllers/schools_controller.rb
invoke test_unit
create test/controllers/schools_controller_test.rb
create app/controllers/programs_controller.rb
invoke test_unit
create test/controllers/programs_controller_test.rb
That's all.
I hope this helps

Generate a controller with all the RESTful functions

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

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