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
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.
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.
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'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
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