如何使继承的资源可选嵌套为单例?
-
13-12-2019 - |
题
我正在使用inherited_resources来干燥我的控制器,但无法弄清楚如何使特定控制器正当行为。在我的模型中,User has_one Person
。我希望它是可选择嵌套的,在嵌套时表现为单身,并且在未嵌套时作为非单例。换句话说,我希望能够列出所有已知的人(/人),获取人员#5(/人/ 5),并获取用户10的唯一人(/用户/ 10 /人)。在路由中如下所示:
resources :users
resource :person
end
resources :people
.
...设置路线,因为我期望:
user_person POST /users/:user_id/person(.:format) people#create
new_user_person GET /users/:user_id/person/new(.:format) people#new
edit_user_person GET /users/:user_id/person/edit(.:format) people#edit
GET /users/:user_id/person(.:format) people#show
PUT /users/:user_id/person(.:format) people#update
DELETE /users/:user_id/person(.:format) people#destroy
people GET /people(.:format) people#index
POST /people(.:format) people#create
new_person GET /people/new(.:format) people#new
edit_person GET /people/:id/edit(.:format) people#edit
person GET /people/:id(.:format) people#show
PUT /people/:id(.:format) people#update
DELETE /people/:id(.:format) people#destroy
.
......如此伟大。现在,如果在peoplecontroller中,我使用:
belongs_to :user, :optional => true
.
然后是非嵌套/人的URL工作,但嵌套/用户/:user_id / person ulls not:undefined method 'people'
如果,而不是peoplecontroller,我使用:
belongs_to :user, :optional => true, :singleton => true
.
然后...然后嵌套/用户/:user_id / person urls工作,但非嵌套/ people网址不会因为它被视为单例,即使是非嵌套:undefined method 'person'
摘要:在通过嵌套路由访问时,是否有一种方法可以使Inherited_resources将资源处理为单例,而是在通过不嵌套路由访问时作为非单例?
解决方案
如果任何人试图做类似的事情,我最终就放弃了Inherited_resources。我发现我的控制器中的“魔法”更幸福。
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