是否可以从现有数据库创建控制器、模型和视图?
我无法通过谷歌搜索找到命令。
这里我说的是逆向工程
您必须为每个具有关系的表创建简单模型,然后您可以
[rails3] > rails generate scaffold_controller Club name:string exclusive:boolean
create app/controllers/clubs_controller.rb
invoke erb
create app/views/clubs
create app/views/clubs/index.html.erb
create app/views/clubs/edit.html.erb
create app/views/clubs/show.html.erb
create app/views/clubs/new.html.erb
create app/views/clubs/_form.html.erb
create app/views/layouts/clubs.html.erb
invoke test_unit
create test/functional/clubs_controller_test.rb
或者,您可以尝试 active_admin gem
活动管理员 -https://github.com/gregbell/active_admin
rails generate active_admin:resource [MyModelName]
RailsAdmin 也足够好https://github.com/sferik/rails_admin
如果您的模型不使用 rails 约定,您应该为它指定至少 2 条规则。例子
class Article < ActiveRecord::Base
self.table_name "tbl_articles"
self.primary_key "art_id"
end
好吧,这违反了原则。如果您想要快速引导您的应用程序,您必须做的越好,那就是复制您在数据库中拥有的模型并使用脚手架。请记住,Rails 使用很多约定,如果您决定不遵循,您将遇到很多麻烦。
如果您需要帮助,请查看本指南。
这就是你可以做到的 -
尝试:
rails g scaffold myscaffold
这将生成文件:
invoke active_record
create db/migrate/20130124100759_create_myscaffolds.rb
create app/models/myscaffold.rb
invoke test_unit
create test/unit/myscaffold_test.rb
create test/fixtures/myscaffolds.yml
route resources :myscaffolds
invoke scaffold_controller
create app/controllers/myscaffolds_controller.rb
invoke erb
create app/views/myscaffolds
create app/views/myscaffolds/index.html.erb
create app/views/myscaffolds/edit.html.erb
create app/views/myscaffolds/show.html.erb
create app/views/myscaffolds/new.html.erb
create app/views/myscaffolds/_form.html.erb
invoke test_unit
create test/functional/myscaffolds_controller_test.rb
invoke helper
create app/helpers/myscaffolds_helper.rb
invoke test_unit
create test/unit/helpers/myscaffolds_helper_test.rb
invoke assets
invoke coffee
create app/assets/javascripts/myscaffolds.js.coffee
invoke scss
create app/assets/stylesheets/myscaffolds.css.scss
invoke scss
identical app/assets/stylesheets/scaffolds.css.scss