16

我正在使用 Ruby 2.0 和 Rails 4.0 构建一个 Ruby on Rails api。我的应用程序几乎完全是一个 JSON API，因此如果发生错误（500、404），我想捕获该错误并返回格式良好的 JSON 错误消息。

我试过这个，而且：

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

在我的应用程序控制器中。

这些都不能解决问题（根本没有捕获异常）。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 中执行此操作的好的文档。

有人知道吗？

编辑 这是我转到 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这么远，应该有东西抓住它并返回适当的 json 错误响应。

4

5 回答 5

17

该请求甚至没有击中您的应用程序。

您需要定义一个包罗万象的路由，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）

修改您的 routes.rb 文件以包含以下内容

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

在你的控制器中

class ErrorsController < ApplicationController

  def catch_404
    raise ActionController::RoutingError.new(params[:path])
  end
end

然后你rescue_from应该抓住错误。

于 2013-12-11T00:47:48.113 回答
16

在尝试了一些变体之后，我认为这是处理 API 404 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
于 2014-08-01T11:50:57.280 回答
9

这适用于 rails4，这样您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
于 2013-09-01T16:35:51.680 回答
2

我使用了公共文件夹中的 404.html，这是在开发环境中。
我实际上得到了答案：

然而，我做了一个小实验，看看是哪些代码真正使它起作用。这是我只添加的代码片段。

配置/路由.rb

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

应用程序/控制器/application_controller.rb

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

将不胜感激任何关于为什么需要一些原件的评论和澄清。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并且raise ActionController::RoutingError似乎是旧 Rails 版本的流行答案。

于 2014-09-09T20:10:49.270 回答
0

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

如果您不希望在您的开发模式下出现这种行为，请在下面添加上面的代码

unless Rails.application.config.consider_all_requests_local

于 2013-08-21T13:48:57.300 回答