Sinatra / Rack`ERROR URI :: InvalidURIError：重定向上的URI（不是URI？）` [英] Sinatra/Rack `ERROR URI::InvalidURIError: bad URI(is not URI?)` on redirect

问题描述

我刚刚开始了我的第一个Sinatra项目，一个简单的电视节目管理网络应用程序，并希望拥有漂亮的网址。所以当一个用户输入搜索框并提交时，我不想要一个 /？search = foobar -style URL，并且要重定向到 /搜索/ foobar的。这也使我能够从主要的 get'/'路由分离 get'/ search /：name 路由。我在过滤器之前使用实现了重定向，并正确地转义了 params [] 变量：

  before do 
 if params.has_key？ 'search'
 redirect to（/ search /＃{URI.escape（params ['search']）}）
 end 
 end 
  

然后我继续

  / search /：query'do 
 result = search_api params [：query] 
 if result =='null'
＃no results 
 else 
 result = JSON。解析（结果）
如果result.key？ '显示'
＃显示搜索结果
 else 
＃重定向到一个单独的显示
＃（result.keys）.first是提供的显示的真实名称
 ＃由api。它可能包含特殊字符
＃
＃（result.keys）.first＃=> 打破坏
＃result.keys＃=> 打破坏
＃result.key？ Breaking Bad＃=> true 
 redirect to（'/ show /＃{URI.escape（（result.keys）.first）}'）
 end 
 end 
 end 
  

不幸的是，重定向到 / show 页面只能工作，如果有除了％之外的名称中没有URI特殊字符。这也意味着没有空格。当我搜索空间或变音符或任何东西的东西，例如 GET for /？search = Breaking％20Bad ，我从Sinatra / Rack得到以下错误：

  [2013-02-02 00:30:29]错误URI :: InvalidURIError：bad URI（is不是URI？）：http：// localhost：9393 / show / Breaking Bad 
 /Users/Ps0ke/.rvm/rubies/ruby-1.9.3-p194/lib/ruby/1.9.1/uri/generic .rb：1202：在rescue in merge
 /Users/Ps0ke/.rvm/rubies/ruby-1.9.3-p194/lib/ruby/1.9.1/uri/generic.rb:1199:in `merge'
 /Users/Ps0ke/.rvm/rubies/ruby-1.9.3-p194/lib/ruby/1.9.1/webrick/httpresponse.rb:220:in`setup_header'
 /用户/ Ps0ke / .rvm / rubies / ruby​​-1.9.3-p194 / lib / ruby​​ / 1.9.1 / webrick / httpresponse.rb：150：在`send_response'
 /Users/Ps0ke/.rvm/rubies $ r $ $ /lib/ruby/1.9.1/webrick/server.rb:191:in`start_thread'中的块
  

浏览器显示我，那是g ot重定向到 / search / Breaking％20Bad ，所以第一个重定向工作。这个错误只发生在搜索产生确切命中时，所以重定向的问题是在 get'/ search /：query'路由中搜索。我记得它工作了一次，但是在我的git历史中找不到正确的提交。

我正在运行

 ％gem list sinatra 
 sinatra（1.3.4,1.3.3）
 
％gem list rack 
机架（1.4.4,1.4.1）
机架缓存（1.2）
 rack-flash3（1.0.3）
机架保护（1.3.2,1.2.0 ）
 rack-ssl（1.3.2）
机架测试（0.6.2，0.6.1）
  

也许有人可以告诉我：

1. 这是好还是坏的做法/还有另一种方法在sinatra到prettifiy URL


2. 如何解决这个问题，因为我认为我已经逃脱了所有的细节

非常感谢你提前：）

解决方案

在行

  redirect to（'/ show /＃{URI.escape（（result.keys）.first）}'）
  pre> 
 
 
您正在使用单引号。这意味着不执行字符串插值，因此文字字符串 / show /＃{URI.escape（（result.keys）.first）} 正被用作url，这就是为什么它是失败的。
 
 
 
为了插值工作，你需要使用这样的双引号：
  redirect to（/ show /＃{URI.escape（（result.keys）.first）}）
  pre> 
 
 
这将导致＃{URI.escape（（result.keys）.first）} 被替换与转义的电影名称，应该是一个有效的URL。
 
 
 
请注意，在您的第一个重定向，你使用双引号，所以它的工作原理：
  redirect to（/ search /＃{URI.escape（params ['search']）}）
  
 
I've just started my first Sinatra project, a simple TV-show management web app, and wanted to have beautiful URLs. So when a user types in the search box and submits, I don't want to have a /?search=foobar-style URL and want do redirect them to /search/foobar. This also enables me to separate the get '/search/:name route from the main get '/' route. I've implemented the redirect using a before filter and properly escaped the params[] variable:
before do
    if params.has_key? 'search'
        redirect to("/search/#{URI.escape(params['search'])}")
    end
end
and later I continue with
get '/search/:query' do
    result = search_api params[:query]
    if result == 'null'
        # no results
    else
        result = JSON.parse(result)
        if result.key? 'shows'
            # display search results
        else
            # redirect to one single show
            # (result.keys).first is the real name of the show provided
            # by the api. It may contain special characters
            #
            # (result.keys).first #=> "Breaking Bad"
            # result.keys #=> "Breaking Bad"
            # result.key? "Breaking Bad" #=> true
            redirect to('/show/#{URI.escape((result.keys).first)}')
        end
    end
end
unfortunately, the redirect to the /show page only works, if there are no URI special characters in the name apart from %. This also means no spaces. When I search for something with a space or an umlaut or anything, e.g. a GET for /?search=Breaking%20Bad, I get the following error from Sinatra/Rack: 
[2013-02-02 00:30:29] ERROR URI::InvalidURIError: bad URI(is not URI?): http://localhost:9393/show/Breaking Bad
    /Users/Ps0ke/.rvm/rubies/ruby-1.9.3-p194/lib/ruby/1.9.1/uri/generic.rb:1202:in `rescue in merge'
    /Users/Ps0ke/.rvm/rubies/ruby-1.9.3-p194/lib/ruby/1.9.1/uri/generic.rb:1199:in `merge'
    /Users/Ps0ke/.rvm/rubies/ruby-1.9.3-p194/lib/ruby/1.9.1/webrick/httpresponse.rb:220:in `setup_header'
    /Users/Ps0ke/.rvm/rubies/ruby-1.9.3-p194/lib/ruby/1.9.1/webrick/httpresponse.rb:150:in `send_response'
    /Users/Ps0ke/.rvm/rubies/ruby-1.9.3-p194/lib/ruby/1.9.1/webrick/httpserver.rb:110:in `run'
    /Users/Ps0ke/.rvm/rubies/ruby-1.9.3-p194/lib/ruby/1.9.1/webrick/server.rb:191:in `block in start_thread'
The browser shows me, that it got redirected to /search/Breaking%20Bad, so the first redirect worked. This "bug" only occurs, when the search produces an exact hit, so the problem with the redirect is to search in the get '/search/:query' route. I remember that it worked once, but can't find the right commit in my git history.

I am running
% gem list sinatra
sinatra (1.3.4, 1.3.3)

% gem list rack
rack (1.4.4, 1.4.1)
rack-cache (1.2)
rack-flash3 (1.0.3)
rack-protection (1.3.2, 1.2.0)
rack-ssl (1.3.2)
rack-test (0.6.2, 0.6.1)
Maybe someone of you can tell me:

whether this is good or bad practices/there is another way in sinatra to prettifiy URLs
how to fix this, since I think I have already escaped everything carefully
Thank you very much in advance :)
解决方案
In the line
redirect to('/show/#{URI.escape((result.keys).first)}')
you are using single quotes. This means that string interpolation isn’t performed, so the literal string /show/#{URI.escape((result.keys).first)} is being used as the url and that is why it is failing.

In order for interpolation to work you need to use double quotes like this:
redirect to("/show/#{URI.escape((result.keys).first)}")
This will cause #{URI.escape((result.keys).first)} to be replaced with the escaped movie name, which should be a valid url.

Note that in your first redirect you do use double quotes, so it works as expected:
redirect to("/search/#{URI.escape(params['search'])}")


                        
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