Rails 4.0的自定义错误处理

Question

我正在使用Ruby 2.0和Rails 4.0构建一个RubyonRails api。我的应用程序几乎只是一个JSON API，所以如果发生错误（500404），我希望捕获该错误并返回一个格式良好的JSON错误消息。

我试过这个，也试过：

rescue_from ActionController::RoutingError, :with => :error_render_method
def error_render_method
  puts "HANDLING ERROR"
  render :json => { :errors => "Method not found." }, :status => :not_found
  true
end

在我的ApplicationController中。

这两者都不起作用（根本没有捕捉到异常）。我的谷歌搜索显示，这在3.1和3.2之间发生了很大变化，我找不到任何关于如何在Rails 4.0中做到这一点的好文档。

有人知道吗？

编辑这是当我转到404页面时的堆栈跟踪：

Started GET "/testing" for 127.0.0.1 at 2013-08-21 09:50:42 -0400
ActionController::RoutingError (No route matches [GET] "/testing"):
actionpack (4.0.0) lib/action_dispatch/middleware/debug_exceptions.rb:21:in `call'
actionpack (4.0.0) lib/action_dispatch/middleware/show_exceptions.rb:30:in `call'
railties (4.0.0) lib/rails/rack/logger.rb:38:in `call_app'
railties (4.0.0) lib/rails/rack/logger.rb:21:in `block in call'
activesupport (4.0.0) lib/active_support/tagged_logging.rb:67:in `block in tagged'
activesupport (4.0.0) lib/active_support/tagged_logging.rb:25:in `tagged'
activesupport (4.0.0) lib/active_support/tagged_logging.rb:67:in `tagged'
railties (4.0.0) lib/rails/rack/logger.rb:21:in `call'
actionpack (4.0.0) lib/action_dispatch/middleware/request_id.rb:21:in `call'
rack (1.5.2) lib/rack/methodoverride.rb:21:in `call'
rack (1.5.2) lib/rack/runtime.rb:17:in `call'
activesupport (4.0.0) lib/active_support/cache/strategy/local_cache.rb:83:in `call'
rack (1.5.2) lib/rack/lock.rb:17:in `call'
actionpack (4.0.0) lib/action_dispatch/middleware/static.rb:64:in `call'
railties (4.0.0) lib/rails/engine.rb:511:in `call'
railties (4.0.0) lib/rails/application.rb:97:in `call'
rack (1.5.2) lib/rack/lock.rb:17:in `call'
rack (1.5.2) lib/rack/content_length.rb:14:in `call'
rack (1.5.2) lib/rack/handler/webrick.rb:60:in `service'
/System/Library/Frameworks/Ruby.framework/Versions/2.0/usr/lib/ruby/2.0.0/webrick/httpserver.rb:138:in `service'
/System/Library/Frameworks/Ruby.framework/Versions/2.0/usr/lib/ruby/2.0.0/webrick/httpserver.rb:94:in `run'
/System/Library/Frameworks/Ruby.framework/Versions/2.0/usr/lib/ruby/2.0.0/webrick/server.rb:295:in `block in start_thread'

Rendered /Library/Ruby/Gems/2.0.0/gems/actionpack-4.0.0/lib/action_dispatch/middleware/templates/rescues/_trace.erb (1.0ms)
Rendered /Library/Ruby/Gems/2.0.0/gems/actionpack-4.0.0/lib/action_dispatch/middleware/templates/routes/_route.html.erb (2.9ms)
Rendered /Library/Ruby/Gems/2.0.0/gems/actionpack-4.0.0/lib/action_dispatch/middleware/templates/routes/_route.html.erb (0.9ms)
Rendered /Library/Ruby/Gems/2.0.0/gems/actionpack-4.0.0/lib/action_dispatch/middleware/templates/routes/_table.html.erb (1.1ms)
Rendered /Library/Ruby/Gems/2.0.0/gems/actionpack-4.0.0/lib/action_dispatch/middleware/templates/rescues/routing_error.erb within rescues/layout (38.3ms)

我认为我不希望它走到这一步，应该有东西抓住它并返回适当的json错误响应。

Answer 1

请求甚至没有到达您的应用程序。

您需要定义一个catchall路由，以便Rails将请求发送到您的应用程序，而不是显示错误（在开发中）或呈现public/404.html页面（在生产中）

修改routes.rb文件以包含以下

match "*path", to: "errors#catch_404", via: :all

在你的控制器

class ErrorsController < ApplicationController
  def catch_404
    raise ActionController::RoutingError.new(params[:path])
  end
end

然后您的rescue_from应该会捕捉到错误。

Answer 2

在尝试了一些变体之后，我决定将其作为处理API 404s:的最简单方法

# Passing request spec
describe 'making a request to an unrecognised path' do
  before { host! 'api.example.com' }
    it 'returns 404' do
    get '/nowhere'
    expect(response.status).to eq(404)
  end
end
# routing
constraints subdomain: 'api' do
  namespace :api, path: '', defaults: { format: 'json' } do
    scope module: :v1, constraints: ApiConstraints.new(1) do
      # ... actual routes omitted ...
    end
    match "*path", to: -> (env) { [404, {}, ['{"error": "not_found"}']] }, via: :all
  end
end

Answer 3

这在rails4中有效，这样您就可以直接管理所有错误：例如，当api调用发生错误时，您可以将error_info呈现为json。。

application_controller.rb

class ApplicationController < ActionController::Base
  protect_from_forgery

  # CUSTOM EXCEPTION HANDLING
  rescue_from StandardError do |e|
    error(e)
  end
  def routing_error
    raise ActionController::RoutingError.new(params[:path])
  end
  protected
  def error(e)
    #render :template => "#{Rails::root}/public/404.html"
    if env["ORIGINAL_FULLPATH"] =~ /^/api/
    error_info = {
      :error => "internal-server-error",
      :exception => "#{e.class.name} : #{e.message}",
    }
    error_info[:trace] = e.backtrace[0,10] if Rails.env.development?
    render :json => error_info.to_json, :status => 500
    else
      #render :text => "500 Internal Server Error", :status => 500 # You can render your own template here
      raise e
    end
  end
  # ...
end

路线.rb

MyApp::Application.routes.draw do
  # ...
  # Any other routes are handled here (as ActionDispatch prevents RoutingError from hitting ApplicationController::rescue_action).
  match "*path", :to => "application#routing_error", :via => :all
end

Answer 4

我使用了来自公共文件夹的404.html，这是在开发环境中
我实际上从得到了答案

• https://stackoverflow.com/posts/20508445/revisions以及
• https://coderwall.com/p/w3ghqq

然而，我做了一个小实验，看看是什么代码片段真正使它发挥了作用。以下是我刚刚添加的代码片段。

config/routes.rb

Rails.application.routes.draw do
    // other routes
    match "*path", to: "application#catch_404", via: :all
end

app/controller/application_controller.rb

class ApplicationController < ActionController::Base
    def catch_404
        render :file => 'public/404.html', :status => :not_found
    end
end

对于为什么需要一些原件，我们将不胜感激。例如，使用这行代码

raise ActionController::RoutingError.new(params[:path])

和这个

rescue_from ActionController::RoutingError, :with => :error_render_method

因为rescue_from和raise ActionController::RoutingError似乎是旧Rails版本中流行的答案。

Answer 5

如果您想以相同的方式响应所有类型的错误，请尝试此操作

rescue_from StandardError, :with => :error_render_method

如果您不希望在开发模式中出现这种行为，请在下添加以上代码

unless Rails.application.config.consider_all_requests_local